Related
I basically wrote a code in which I take two command line arguments one being the type of file that I want to search in my directory and they other being the amount I want(which is not implemented yet, but I can fix that)
The code is like so:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <unistd.h>
#define sizeFileName 500
#define filesMax 5000
int cmpfunc( const void *a, const void *b) {
return *(char*)a + *(char*)b;
}
int main( int argc, char ** argv) {
FILE * fp = popen( "find . -type f", "r");
char * type = argv[1];
char * extension = ".";
char* tExtension;
tExtension = malloc(strlen(type)+1+4);
strcpy(tExtension, extension);
strcat(tExtension, type);
// printf("%s\n",tExtension);
int amount = atoi(argv[2]);
//printf("%d\n",amount);
char buff[sizeFileName];
int nFiles = 0;
char * files[filesMax];
while(fgets(buff,sizeFileName,fp)) {
int leng = strlen(buff) - 1;
if (strncmp(buff + leng - 4, tExtension, 4) == 0){
files[nFiles] = strndup(buff,leng);
//printf("\t%s\n", files[nFiles]);
nFiles ++;
}
}
fclose(fp);
printf("Found %d files\n", nFiles);
long long totalBytes = 0;
struct stat st;
// sorting based on byte size from greatest to least
qsort(files, (size_t) strlen(files), (size_t) sizeof(char), cmpfunc);
for(int i = 0;i< nFiles; i ++) {
if(0!= stat(files[i],&st)){
perror("stat failed:");
exit(-1);
}
totalBytes += st.st_size;
printf("%s : %ld\n",files[i],st.st_size);
}
printf("Total size: %lld\n", totalBytes);
// clean up
for(int i = 0; i < nFiles ; i ++ ) {
free(files[i]);
}
return 0;
}
So far I have every section set up properly, upon running the code say $./find ini 5, it would print out all the ini files followed by their byte size(it's currently ignore the 5). However, for the qsort(), I'm not exactly sure how I would sort the contents of char * files as while it holds the pathnames, I had to use stat to get the byte sizes, how would I print out a sorted version of my print statements featuring the first statement being the most bytes and finishes at the least bytes?
If we suppose your input is valid, your question could be simplified with:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define filesMax 5000
int cmpfunc(const void const *a, const void *b) { return *(char *)a + *(char *)b; }
int main(void) {
int nFiles = 4;
char *files[filesMax] = {"amazing", "hello", "this is a file", "I'm a bad file"};
qsort(files, strlen(files), sizeof(char), cmpfunc);
for (int i = 0; i < nFiles;; i++) {
printf("%s\n", files[i]);
}
}
If you compile with warning that give you:
source_file.c:11:23: warning: incompatible pointer types passing 'char *[5000]' to parameter of type 'const char *' [-Wincompatible-pointer-types]
qsort(files, strlen(files), sizeof(char), cmpfunc);
^~~~~
qsort() expect the size of your array (or in your case a subsize) and it's also expect the size of one element of your array. In both you wrongly give it to it. Also, your compare function doesn't compare anything, you are currently adding the first bytes of both pointer of char, that doesn't make a lot of sense.
To fix your code you must write:
qsort(files, nFiles, sizeof *files, &cmpfunc);
and also fix your compare function:
int cmpfunc_aux(char * const *a, char * const *b) { return strcmp(*a, *b); }
int cmpfunc(void const *a, void const *b) { return cmpfunc_aux(a, b); }
also size should be of type size_t:
size_t nFiles = 0;
Don't forget that all informations about how to use a function are write in their doc.
how would I print out a sorted version of my print statements featuring the first statement being the most bytes and finishes at the least bytes?
Your code don't show any clue that your are trying to do that, you are currently storing name file and only that. How do you expect sort your file with an information you didn't acquired ?
However, that simple create a struct that contain both file name and size, acquire information needed to sort it and sort it:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/stat.h>
#include <inttypes.h>
struct file {
off_t size;
char *name;
};
int cmpfunc_aux(struct file const *a, struct file const *b) {
if (a->size > b->size) {
return -1;
} else if (a->size < b->size) {
return 1;
} else {
return 0;
}
}
int cmpfunc(void const *a, void const *b) { return cmpfunc_aux(a, b); }
#define filesMax 5000
int main(void) {
size_t nFiles = 4;
struct file files[filesMax] = {{42, "amazing"},
{21, "hello"},
{168, "this is a file"},
{84, "I'm a bad file"}};
qsort(files, nFiles, sizeof *files, &cmpfunc);
for (size_t i = 0; i < nFiles; i++) {
printf("%s, %" PRId64 "\n", files[i].name, (intmax_t)files[i].size);
}
}
The function cmpfunc() provided adds the first character of each string, and that's not a proper comparison function (it should give a opposite sign value when you switch the parameters, e.g. if "a" and "b" are the strings to compare, it adds the first two characters of both strings, giving 97+98 == 195, which is positive on unsigned chars, then calling with "b" and "a" should give a negative number (and it again gives you 98 + 97 == 195), more on, it always gives the same result ---even with signed chars--- so it cannot be used as a sorting comparator)
As you are comparing strings, why not to use the standard library function strcmp(3) which is a valid comparison function? It gives a negative number if first string is less lexicographically than the second, 0 if both are equal, and positive if first is greater lexicographically than the second.
if your function has to check (and sort) by the lenght of the filenames, then you can define it as:
int cmpfunc(char *a, char *b) /* yes, you can define parameters as char * */
{
return strlen(a) - strlen(b);
}
or, first based on file length, then lexicographically:
int cmpfunc(char *a, char *b)
{
int la = strlen(a), lb = strlen(b);
if (la != lb) return la - lb;
/* la == lb, so we must check lexicographycally */
return strcmp(a, b);
}
Now, to continue helping you, I need to know why do you need to sort anything, as you say that you want to search a directory for a file, where does the sorting take place in the problem?
I use an Arduino Uno with Arduino IDE 1.8.3. I have two arrays. I want to write a Deputy function that can add two arrays, and return the result to the main function and print it.
But I want to use x(sizeof(a)), but it seems not correct...
How do I solve this problem?
This is my code:
int a[]={1,2,3,4,5,6},b[]={1,1,1,1,1,1};
void setup() {
Serial.begin(9600);
int *p;
p = add(a,b);
for(int i=0;i<4;i++){
Serial.print(*(p+i));
}
}
void loop() {
}
int * add(int *a,int *b) {
int x = sizeof(a);
int y = sizeof(b);
static int z[4];
for(int i=0;i<4;i++) {
z[i]=a[i]+b[i];
}
return z;
}
int* a does not know the size of the array.
Easiest pass it as an extra parameter.
The next problem is that your static result cannot change its size dynamically.
static has additional problems anyway, in general.
int* add(const int *a,const int *b, int* result, byte size) {
for(byte i=0; i<size; i++) {
result[i]=a[i]+b[i];
}
return result;
}
Returning the result as the return value may be convenient.
This question already has an answer here:
Summing the rows of a matrix (stored in either row-major or column-major order) in CUDA
(1 answer)
Closed 5 years ago.
I declared two GPU memory pointers, and allocated the GPU memory, transfer data and launch the kernel in the main:
// declare GPU memory pointers
char * gpuIn;
char * gpuOut;
// allocate GPU memory
cudaMalloc(&gpuIn, ARRAY_BYTES);
cudaMalloc(&gpuOut, ARRAY_BYTES);
// transfer the array to the GPU
cudaMemcpy(gpuIn, currIn, ARRAY_BYTES, cudaMemcpyHostToDevice);
// launch the kernel
role<<<dim3(1),dim3(40,20)>>>(gpuOut, gpuIn);
// copy back the result array to the CPU
cudaMemcpy(currOut, gpuOut, ARRAY_BYTES, cudaMemcpyDeviceToHost);
cudaFree(gpuIn);
cudaFree(gpuOut);
And this is my code inside the kernel:
__global__ void role(char * gpuOut, char * gpuIn){
int idx = threadIdx.x;
int idy = threadIdx.y;
char live = '0';
char dead = '.';
char f = gpuIn[idx][idy];
if(f==live){
gpuOut[idx][idy]=dead;
}
else{
gpuOut[idx][idy]=live;
}
}
But here are some errors, I think here are some errors on the pointers. Any body can give a help?
The key concept is the storage order of multidimensional arrays in memory -- this is well described here. A useful abstraction is to define a simple class which encapsulates a pointer to a multidimensional array stored in linear memory and provides an operator which gives something like the usual a[i][j] style access. Your code could be modified something like this:
template<typename T>
struct array2d
{
T* p;
size_t lda;
__device__ __host__
array2d(T* _p, size_t _lda) : p(_p), lda(_lda) {};
__device__ __host__
T& operator()(size_t i, size_t j) {
return p[j + i * lda];
}
__device__ __host__
const T& operator()(size_t i, size_t j) const {
return p[j + i * lda];
}
};
__global__ void role(array2d<char> gpuOut, array2d<char> gpuIn){
int idx = threadIdx.x;
int idy = threadIdx.y;
char live = '0';
char dead = '.';
char f = gpuIn(idx,idy);
if(f==live){
gpuOut(idx,idy)=dead;
}
else{
gpuOut(idx,idy)=live;
}
}
int main()
{
const int rows = 5, cols = 6;
const size_t ARRAY_BYTES = sizeof(char) * size_t(rows * cols);
// declare GPU memory pointers
char * gpuIn;
char * gpuOut;
char currIn[rows][cols], currOut[rows][cols];
// allocate GPU memory
cudaMalloc(&gpuIn, ARRAY_BYTES);
cudaMalloc(&gpuOut, ARRAY_BYTES);
// transfer the array to the GPU
cudaMemcpy(gpuIn, currIn, ARRAY_BYTES, cudaMemcpyHostToDevice);
// launch the kernel
role<<<dim3(1),dim3(rows,cols)>>>(array2d<char>(gpuOut, cols), array2d<char>(gpuIn, cols));
// copy back the result array to the CPU
cudaMemcpy(currOut, gpuOut, ARRAY_BYTES, cudaMemcpyDeviceToHost);
cudaFree(gpuIn);
cudaFree(gpuOut);
return 0;
}
The important point here is that a two dimensional C or C++ array stored in linear memory can be addressed as col + row * number of cols. The class in the code above is just a convenient way of expressing this.
First off, here is some code:
int main()
{
int days[] = {1,2,3,4,5};
int *ptr = days;
printf("%u\n", sizeof(days));
printf("%u\n", sizeof(ptr));
return 0;
}
Is there a way to find out the size of the array that ptr is pointing to (instead of just giving its size, which is four bytes on a 32-bit system)?
No, you can't. The compiler doesn't know what the pointer is pointing to. There are tricks, like ending the array with a known out-of-band value and then counting the size up until that value, but that's not using sizeof().
Another trick is the one mentioned by Zan, which is to stash the size somewhere. For example, if you're dynamically allocating the array, allocate a block one int bigger than the one you need, stash the size in the first int, and return ptr+1 as the pointer to the array. When you need the size, decrement the pointer and peek at the stashed value. Just remember to free the whole block starting from the beginning, and not just the array.
The answer is, "No."
What C programmers do is store the size of the array somewhere. It can be part of a structure, or the programmer can cheat a bit and malloc() more memory than requested in order to store a length value before the start of the array.
For dynamic arrays (malloc or C++ new) you need to store the size of the array as mentioned by others or perhaps build an array manager structure which handles add, remove, count, etc. Unfortunately C doesn't do this nearly as well as C++ since you basically have to build it for each different array type you are storing which is cumbersome if you have multiple types of arrays that you need to manage.
For static arrays, such as the one in your example, there is a common macro used to get the size, but it is not recommended as it does not check if the parameter is really a static array. The macro is used in real code though, e.g. in the Linux kernel headers although it may be slightly different than the one below:
#if !defined(ARRAY_SIZE)
#define ARRAY_SIZE(x) (sizeof((x)) / sizeof((x)[0]))
#endif
int main()
{
int days[] = {1,2,3,4,5};
int *ptr = days;
printf("%u\n", ARRAY_SIZE(days));
printf("%u\n", sizeof(ptr));
return 0;
}
You can google for reasons to be wary of macros like this. Be careful.
If possible, the C++ stdlib such as vector which is much safer and easier to use.
There is a clean solution with C++ templates, without using sizeof(). The following getSize() function returns the size of any static array:
#include <cstddef>
template<typename T, size_t SIZE>
size_t getSize(T (&)[SIZE]) {
return SIZE;
}
Here is an example with a foo_t structure:
#include <cstddef>
template<typename T, size_t SIZE>
size_t getSize(T (&)[SIZE]) {
return SIZE;
}
struct foo_t {
int ball;
};
int main()
{
foo_t foos3[] = {{1},{2},{3}};
foo_t foos5[] = {{1},{2},{3},{4},{5}};
printf("%u\n", getSize(foos3));
printf("%u\n", getSize(foos5));
return 0;
}
Output:
3
5
As all the correct answers have stated, you cannot get this information from the decayed pointer value of the array alone. If the decayed pointer is the argument received by the function, then the size of the originating array has to be provided in some other way for the function to come to know that size.
Here's a suggestion different from what has been provided thus far,that will work: Pass a pointer to the array instead. This suggestion is similar to the C++ style suggestions, except that C does not support templates or references:
#define ARRAY_SZ 10
void foo (int (*arr)[ARRAY_SZ]) {
printf("%u\n", (unsigned)sizeof(*arr)/sizeof(**arr));
}
But, this suggestion is kind of silly for your problem, since the function is defined to know exactly the size of the array that is passed in (hence, there is little need to use sizeof at all on the array). What it does do, though, is offer some type safety. It will prohibit you from passing in an array of an unwanted size.
int x[20];
int y[10];
foo(&x); /* error */
foo(&y); /* ok */
If the function is supposed to be able to operate on any size of array, then you will have to provide the size to the function as additional information.
For this specific example, yes, there is, IF you use typedefs (see below). Of course, if you do it this way, you're just as well off to use SIZEOF_DAYS, since you know what the pointer is pointing to.
If you have a (void *) pointer, as is returned by malloc() or the like, then, no, there is no way to determine what data structure the pointer is pointing to and thus, no way to determine its size.
#include <stdio.h>
#define NUM_DAYS 5
typedef int days_t[ NUM_DAYS ];
#define SIZEOF_DAYS ( sizeof( days_t ) )
int main() {
days_t days;
days_t *ptr = &days;
printf( "SIZEOF_DAYS: %u\n", SIZEOF_DAYS );
printf( "sizeof(days): %u\n", sizeof(days) );
printf( "sizeof(*ptr): %u\n", sizeof(*ptr) );
printf( "sizeof(ptr): %u\n", sizeof(ptr) );
return 0;
}
Output:
SIZEOF_DAYS: 20
sizeof(days): 20
sizeof(*ptr): 20
sizeof(ptr): 4
There is no magic solution. C is not a reflective language. Objects don't automatically know what they are.
But you have many choices:
Obviously, add a parameter
Wrap the call in a macro and automatically add a parameter
Use a more complex object. Define a structure which contains the dynamic array and also the size of the array. Then, pass the address of the structure.
You can do something like this:
int days[] = { /*length:*/5, /*values:*/ 1,2,3,4,5 };
int *ptr = days + 1;
printf("array length: %u\n", ptr[-1]);
return 0;
My solution to this problem is to save the length of the array into a struct Array as a meta-information about the array.
#include <stdio.h>
#include <stdlib.h>
struct Array
{
int length;
double *array;
};
typedef struct Array Array;
Array* NewArray(int length)
{
/* Allocate the memory for the struct Array */
Array *newArray = (Array*) malloc(sizeof(Array));
/* Insert only non-negative length's*/
newArray->length = (length > 0) ? length : 0;
newArray->array = (double*) malloc(length*sizeof(double));
return newArray;
}
void SetArray(Array *structure,int length,double* array)
{
structure->length = length;
structure->array = array;
}
void PrintArray(Array *structure)
{
if(structure->length > 0)
{
int i;
printf("length: %d\n", structure->length);
for (i = 0; i < structure->length; i++)
printf("%g\n", structure->array[i]);
}
else
printf("Empty Array. Length 0\n");
}
int main()
{
int i;
Array *negativeTest, *days = NewArray(5);
double moreDays[] = {1,2,3,4,5,6,7,8,9,10};
for (i = 0; i < days->length; i++)
days->array[i] = i+1;
PrintArray(days);
SetArray(days,10,moreDays);
PrintArray(days);
negativeTest = NewArray(-5);
PrintArray(negativeTest);
return 0;
}
But you have to care about set the right length of the array you want to store, because the is no way to check this length, like our friends massively explained.
This is how I personally do it in my code. I like to keep it as simple as possible while still able to get values that I need.
typedef struct intArr {
int size;
int* arr;
} intArr_t;
int main() {
intArr_t arr;
arr.size = 6;
arr.arr = (int*)malloc(sizeof(int) * arr.size);
for (size_t i = 0; i < arr.size; i++) {
arr.arr[i] = i * 10;
}
return 0;
}
No, you can't use sizeof(ptr) to find the size of array ptr is pointing to.
Though allocating extra memory(more than the size of array) will be helpful if you want to store the length in extra space.
int main()
{
int days[] = {1,2,3,4,5};
int *ptr = days;
printf("%u\n", sizeof(days));
printf("%u\n", sizeof(ptr));
return 0;
}
Size of days[] is 20 which is no of elements * size of it's data type.
While the size of pointer is 4 no matter what it is pointing to.
Because a pointer points to other element by storing it's address.
In strings there is a '\0' character at the end so the length of the string can be gotten using functions like strlen. The problem with an integer array, for example, is that you can't use any value as an end value so one possible solution is to address the array and use as an end value the NULL pointer.
#include <stdio.h>
/* the following function will produce the warning:
* ‘sizeof’ on array function parameter ‘a’ will
* return size of ‘int *’ [-Wsizeof-array-argument]
*/
void foo( int a[] )
{
printf( "%lu\n", sizeof a );
}
/* so we have to implement something else one possible
* idea is to use the NULL pointer as a control value
* the same way '\0' is used in strings but this way
* the pointer passed to a function should address pointers
* so the actual implementation of an array type will
* be a pointer to pointer
*/
typedef char * type_t; /* line 18 */
typedef type_t ** array_t;
int main( void )
{
array_t initialize( int, ... );
/* initialize an array with four values "foo", "bar", "baz", "foobar"
* if one wants to use integers rather than strings than in the typedef
* declaration at line 18 the char * type should be changed with int
* and in the format used for printing the array values
* at line 45 and 51 "%s" should be changed with "%i"
*/
array_t array = initialize( 4, "foo", "bar", "baz", "foobar" );
int size( array_t );
/* print array size */
printf( "size %i:\n", size( array ));
void aprint( char *, array_t );
/* print array values */
aprint( "%s\n", array ); /* line 45 */
type_t getval( array_t, int );
/* print an indexed value */
int i = 2;
type_t val = getval( array, i );
printf( "%i: %s\n", i, val ); /* line 51 */
void delete( array_t );
/* free some space */
delete( array );
return 0;
}
/* the output of the program should be:
* size 4:
* foo
* bar
* baz
* foobar
* 2: baz
*/
#include <stdarg.h>
#include <stdlib.h>
array_t initialize( int n, ... )
{
/* here we store the array values */
type_t *v = (type_t *) malloc( sizeof( type_t ) * n );
va_list ap;
va_start( ap, n );
int j;
for ( j = 0; j < n; j++ )
v[j] = va_arg( ap, type_t );
va_end( ap );
/* the actual array will hold the addresses of those
* values plus a NULL pointer
*/
array_t a = (array_t) malloc( sizeof( type_t *) * ( n + 1 ));
a[n] = NULL;
for ( j = 0; j < n; j++ )
a[j] = v + j;
return a;
}
int size( array_t a )
{
int n = 0;
while ( *a++ != NULL )
n++;
return n;
}
void aprint( char *fmt, array_t a )
{
while ( *a != NULL )
printf( fmt, **a++ );
}
type_t getval( array_t a, int i )
{
return *a[i];
}
void delete( array_t a )
{
free( *a );
free( a );
}
#include <stdio.h>
#include <string.h>
#include <stddef.h>
#include <stdlib.h>
#define array(type) struct { size_t size; type elem[0]; }
void *array_new(int esize, int ecnt)
{
size_t *a = (size_t *)malloc(esize*ecnt+sizeof(size_t));
if (a) *a = ecnt;
return a;
}
#define array_new(type, count) array_new(sizeof(type),count)
#define array_delete free
#define array_foreach(type, e, arr) \
for (type *e = (arr)->elem; e < (arr)->size + (arr)->elem; ++e)
int main(int argc, char const *argv[])
{
array(int) *iarr = array_new(int, 10);
array(float) *farr = array_new(float, 10);
array(double) *darr = array_new(double, 10);
array(char) *carr = array_new(char, 11);
for (int i = 0; i < iarr->size; ++i) {
iarr->elem[i] = i;
farr->elem[i] = i*1.0f;
darr->elem[i] = i*1.0;
carr->elem[i] = i+'0';
}
array_foreach(int, e, iarr) {
printf("%d ", *e);
}
array_foreach(float, e, farr) {
printf("%.0f ", *e);
}
array_foreach(double, e, darr) {
printf("%.0lf ", *e);
}
carr->elem[carr->size-1] = '\0';
printf("%s\n", carr->elem);
return 0;
}
#define array_size 10
struct {
int16 size;
int16 array[array_size];
int16 property1[(array_size/16)+1]
int16 property2[(array_size/16)+1]
} array1 = {array_size, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
#undef array_size
array_size is passing to the size variable:
#define array_size 30
struct {
int16 size;
int16 array[array_size];
int16 property1[(array_size/16)+1]
int16 property2[(array_size/16)+1]
} array2 = {array_size};
#undef array_size
Usage is:
void main() {
int16 size = array1.size;
for (int i=0; i!=size; i++) {
array1.array[i] *= 2;
}
}
Most implementations will have a function that tells you the reserved size for objects allocated with malloc() or calloc(), for example GNU has malloc_usable_size()
However, this will return the size of the reversed block, which can be larger than the value given to malloc()/realloc().
There is a popular macro, which you can define for finding number of elements in the array (Microsoft CRT even provides it OOB with name _countof):
#define countof(x) (sizeof(x)/sizeof((x)[0]))
Then you can write:
int my_array[] = { ... some elements ... };
printf("%zu", countof(my_array)); // 'z' is correct type specifier for size_t
so i know the bases of programming, i have a decent amount of experience with java, but im learning C for school right now. I still dont completely understand the whole pointer aspect, which is what im sure caused the fault. This program works fine when run on my computer, but when i try and run it on my schools unix shell it gives me a seg fault. if someone could please explain to me why or how ive misused hte pointers, that would help me greatly.
//Matthew Gerton
//CS 222 - 002
//10/10/14
//HW Six
//libraries
#include<stdio.h>
#include<string.h>
#define max_Length 256
//prototypes
void decode(char *a, char *b);
void trimWhite(char *a);
void encode(char *a, char *b);
int main(void)
{
//character arrays
char coded[max_Length], decoded[max_Length];
//decode the sample phrase
char sample[] = {'P','H','H','W','D','W','C','R','R','F','D','Q','F','H','O','H','G','J',
'R','W','R','P','H','W','U','R','K','R','W','H','O','U','R','R','P','I','R','X','U'};
decode(sample, decoded);
//scans a user input string to decode, and decodes it
printf("\nPlease enter a phrase to decode: ");
gets(coded);
trimWhite(coded);
decode(coded, decoded);
//scans a user input phrase to encode
printf("\nPlease enter a phrase to encode: ");
gets(coded);
trimWhite(coded);
encode(coded, decoded);
}
//removes any spaces from the input
void trimWhite(char *a)
{
char temp[max_Length];
int z=0, y=0;
while(a[z]!='\0')
{
if(a[z]!=' ')
{
temp[y]=a[z];
y++;
}
z++;
}
temp[y] = '\0';
strcpy(a,temp);
}
//decodes any phrase
void decode(char *a, char *b)
{
int i=0,n;
memset(b, '\0', sizeof(b));
while(a[i]!='\0')
{
n=(int)a[i];
if(n<97)
n=n+32;
if(n<=99)
n=n+23;
else
n = n-3;
b[i]= (char) n;
i++;
}
b[i]='\0';
printf("Coded message: %s\n", a);
printf("Decoded message: %s\n", b);
}
//codes an input phrase
void encode(char *a, char *b)
{
int i=0,n;
memset(b, '\0', sizeof(b));
strcpy(b,a);
while(a[i]!='\0')
{
n=(int)a[i];
if(n<97)
a[i] = (char)(n+32);
if((n>120)
a[i] = (char)(n-23);
else
a[i] = (char)((n+3);
i++;
}
printf("Coded message: %s\n", a);
}
Your main problem is here:
char sample[] = {'P','H','H', /* snip */ ,'R','X','U'};
The sample[] array is not zero-terminated which may cause the decode() function to copy many more characters than intended, thus overwriting other variables. You need to explicitly add a terminating zero when using an initializer-list:
char sample[] = {'P','H','H', /* ... */ ,'R','X','U',0};
Or you can initialize the array using a string literal, which does include a terminating zero:
char sample[] = "PHHWDWCRRFDQFHOHGJRWRPHWURKRWHOURRPIRXU";
You should probably read "Why is the gets function dangerous".
...
void decode(char *a, char *b)
{
int i=0,n;
memset(b, '\0', sizeof(b));
Also note that the size of the array is lost when it is passed to a function. The function only receives a pointer to its first element. The memset() call above will only zero sizeof(char*) bytes (usually 4 or 8). This doesn't matter though because, as far as I can tell, you only need to zero the first byte. You could simply write:
b[0] = 0;