Hello brothers I have work for along time to show the first 6 character from double value so I couldn't do that and I need a simple method, Ex:
Here's my double value I need to cast it: 14.7534343267653.
I need to show the first 7 characters only: 14.7534.
So any body can solve this with a simple way.
The example of doing what you are looking for in C:
#include <stdio.h>
int main(int argc, char *argv[]) {
double number = 14.7534343267653;
printf("%2.4f", number);
return 0;
}
Other languages have identical or very similar format specifiers, so esentially "%2.4f" and some string format() might be what you are looking for.
Related
I have an empty program:
int main(int argc, char **argv)
{
return 0;
}
When I run nm path/to/exe --format=sysv I get numbers in the symbol names
e.g file_mtx_85 and free_fn_89.
With different program I can get duplicates like var_101 and var_102.
I think the duplication is because I have more than one symbol with that name, but why do I get numbers in the symbol names and what do they mean?
I'm compiling with Xtensa compiler.
I have been searching about this for so long, but i am not able to understand what this question means.
Question:
Write a program in any language to determine how your computer handles graceful
underflow.
I understand that a overflow condition is something like this:
if an integer can store a maximum value of x and if we assign a value of x+1, the value x+1 will be converted to the the lowest value the integer can hold. I understand that underflow is just the reverse.
How does it stand from High performance scientific computing / Linear algebra point of view ?
I have read this link , but i think it's the same underflow/ overflow stuff that i mentioned above. What does the graceful underflow stand for?
Okay,as per the link posted by #StoneBird in this link was particularly helpful. Here i have created a program in c that demonstrates the same.
#include <stdio.h>
#include <math.h>
int main(int argc, char **argv)
{
unsigned int s,e,m;
unsigned int* ptr;
float a,temp=0;
a=1;
float min=pow(2,-129);
while(a>min){
temp=a;
a=a/2;
}
printf("Value=%e\n",temp);
ptr=(unsigned int*)&temp;
s = *ptr >> 31;
e = *ptr & 0x7f800000;
e >>= 23;
m = *ptr & 0x07fffff;
printf("sign = %x\n",s);
printf("exponent = %x\n",e);
printf("mantissa = %x\n",m);
return 0;
}
Here the min variable is used to change the final number...i used min=pow(2,-129), pow(2,-128) and pow(2,-130) to see the results and the saw the Denormal number appear.This wiki page explains it all.
I'm trying to convert raw hex/binary data to different file types.
#include <QByteArray>
#include <QDebug>
int main(int argc, char *argv[])
{
QByteArray package;
package.append( QByteArray::fromHex("a1"));
// "a1" is what is written to the memory, not the string representation of "a1"
qDebug() << package.toHex(); // "a1"
qDebug() << package; // "�"
qDebug() << package.toInt(); // 0
}
Why is the int representation 0 and not 161?
toInt has totally different purpose. It parses string representation of integer. If you want integer representing the value of the first byte of the array, use package[0]. It has char type. I don't remember how qDebug() represents char type, but if you have any problems with it, just static_cast it to unsigned int.
QByteArray::toInt expects that QByteArray contains a string of characters (in ASCII probably), not the binary representation of the number.
If you want to convert binary representation to integer you can use reinterpret_cast:
int i = *reinterpret_cast<quint8*>(package.constData());
Or better use qFromBigEndian/qFromLittleEndian:
int i = qFromLittleEndian<quint8>((const uchar*)package.constData())
In both cases you must know exactly in what format the number is stored and use proper type and endianness.
i'm trying to write a program that gets from the user a value x and prints sinx using taylor series. but my output is bad. the output i get is not even a number, its -1.#IND00 regardless of what i input.
here's my code
#include <stdio.h>
#include <conio.h>
void main()
{
int i;
double x,sum,last;
sum=(double)0;
scanf("%f",&x);
last=x;
sum=last;
for(i=1;i<10;i++)
{
last*=(double)(-x*x)/((2*i)*(2*i+1));
sum+=last;
}
printf("%f",sum);
getch();
}
I can see one problem:
scanf("%f",&x);
x is a double, so you need the l, i.e. "%lf".
[true but irrelevant point about how this isn't the right formula for sinh, even though sinh is nowhere mentioned in the question, redacted..]
I have a global variable that is a *char. My main function header reads as int main(int argc, char* argv[argc]){...}. These two lines of code have to remain the way they are. The first argument of my main function is a number of type *char, that I convert to a char using atoi(...);. I am basically changing the ASCII value to its corresponding character. Now I want to store this local variable character I have into the global variable that is a char pointer. I know the problem is related to allocation of memory, but I am not sure how to go about this.
My code:
char* delim;
int main(int argc, char* argv[argc])
{
char delimCharacter;
if (isdigit(*(argv[3])) == 0) delim = argv[3]; //you can pass in a character or its ascii value
else { //if the argument is a number, then the ascii value is taken
delimCharacter = atoi((argv[3]));
printf("%s\t,%c,\n", argv[3], delimCharacter);
//sprintf( delim, "%c", delimCharacter ); // a failed attempt to do this
*delim = delimCharacter;
//strncpy(delim, delimCharacter, 1); // another failed attempt to do this
}
//printf("%s\n",delim);
This yields a seg fault.
You need to verify you have got (at least) 3 arguments before you start using them.
if (argc < 4)
{
printf("Need 3 args");
exit(1);
}
Then you need to allocate some memory to put the character in.
delim = malloc(2);
// TODO: Should check the result of malloc before using it.
*delim = delimCharacter;
delim[1] = 0; // Need to NULL terminate char*
You're dereferencing an uninitialized pointer. delim never gets initialized when it goes into the else block.
char delim[] = ","; // anything really, as long as as it's one character string
...
delim[0] = delimCharacter;
In addition to your memory issue, I think you are confused about what atoi does. It parses a string representation of a number and returns the equivalent int value, e.g. "10000" => 10,000. I think that you think it will give you the ASCII value of a character, e.g. "A" =>65.
Since you have a char *, and you are (I think) assuming that it contains a single character, you could simply do this:
delimCharacter = *(argv[3]);
However, there really seems to be no need to use the intermediate step of assigning this value to a char variable at all. If the end goal is to have delim point to the char that is the delimiter, then it seems this is all you need to do:
delim = argv[3];
Not only does this remove unnecessary code, but it means you would no longer need to allocate additional memory for delim to point to.
I would also declare delim as a const char * since I assume there is no reason to change it.