I'm working for my classes on a BNF exercise. We must explain how to declare a float and an integer based on this (imaginary?) language :
PROGRAM Z
FLOAT R; I;
FLOAT P;
INT A;
INT B; C; D;
BEGIN
R = 3;2;
I = R + 1,5;
A = 5;
B = 5 * P;
END
The name of this language isn't anywhere in the question (it's in French so I can't copy it in here), but I don't understand the use of ";" here.
I guess it declares R and I at the same time, and then P, but I don't understand the "R=3;2;" line. If a ";" mark a break, the program is just executing 2? Or do we attribute "3" and then "2"? Or is it just a syntax error since it's suppose to be a float and the real thing is supposed to be "3,2"?
What do you think ?
Related
I'm learning C++, and encountering these problems in a simple program, so please help me out.
This is the code
#include<iostream>
using std::cout;
int main()
{ float pie;
pie = (22/7);
cout<<"The Value of Pi(22/7) is "<< pie<<"\n";
return 0;
}
and the output is
The Value of Pi(22/7) is 3
Why is the value of Pi not in decimal?
That's because you're doing integer division.
What you want is really float division:
#include<iostream>
using std::cout;
int main()
{
float pie;
pie = float(22)/7;// 22/(float(7)) is also equivalent
cout<<"The Value of Pi(22/7) is "<< pie<<"\n";
return 0;
}
However, this type conversion: float(variable) or float(value) isn't type safe.
You could have gotten the value you wanted by ensuring that the values you were computing were floating point to begin with as follows:
22.0/7
OR
22/7.0
OR
22.0/7.0
But, that's generally a hassle and will involve that you keep track of all the types you're working with. Thus, the final and best method involves using static_cast:
static_cast<float>(22)/7
OR
22/static_cast<float>(7)
As for why you should use static_cast - see this:
Why use static_cast<int>(x) instead of (int)x?
pie = (22/7);
Here the division is integer division, because both operands are int.
What you intend to do is floating-point division:
pie = (22.0/7);
Here 22.0 is double, so the division becomes floating-point division (even though 7 is still int).
The rule is that IF both operands are integral type (such as int, long, char etc), then it is integer division, ELSE it is floating-point division (i.e when even if a single operand is float or double).
Use:
pi = 22/7.0
If u give the two operands to the / operator as integer then the division performed will be integer division and a float will not be the result.
There is an example of shuffle of OpenCL during the document.
//Examples that are not valid are:
uint8 mask;
short16 a;
short8 b;
b = shuffle(a, mask); // invalid
But I can not understand why. I test this during Android with AndroidStudio, and the result said:build program failed:BC-src-code:9:9:{9:9-9:16}: error: no matching builtin function for call to 'shuffle'. Then, I change the short to int, like this:
uint8 mask;
int16 a;
int8 b;
b = shuffle(a, mask);
and it is ok. I can not find any reason from the document, can anybody help me?
Thanks!
I think the critical part of the description in the spec is this:
The size of each element in the mask must match the size of each element in the result.
I take that to mean that if you want to shuffle a vector of shorts, your mask must be a vector of ushort; a mask of uint8 would only be valid for shuffling vectors with elements of 4 bytes - in other words, int, uint, and float.
So the following should be valid again:
ushort8 mask; // <-- changed
short16 a;
short8 b;
b = shuffle(a, mask); // now valid
I need to copy one instance of a structure to another, using the respective structure pointers. The code I have tried is as follows:
typedef struct{
int a, b, c;} test;
int main(){
test *q, *w;
(*w).a = 2;
(*w).b = 3;
(*w).c = 4;
printf("\n%d\n%d\n%d", (*w).a, (*w).b, (*w).c);
memcpy((void*)q, (void*)w, sizeof(test));
printf("\n%d\n%d\n%d", (*q).a, (*q).b, (*q).c);
return 0;
The output I get is:
2
3
4
1875984
32768
1296528
Can someone please tell me how to copy the structure? I need to use pointers to structures, simply doing:
test w, q;
q = w;
will not suffice for my program.
Thanks.
Replace the line in your code:
memcpy((void*)q, (void*)w, sizeof(test));
with the following line:
memcpy((void*)&q, (void*)&w, sizeof(test));
OK, say I have a boolean array called bits, and an int called cursor
I know I can access individual bits by using bits[cursor], and that I can use bit logic to get larger datatypes from bits, for example:
short result = (bits[cursor] << 3) |
(bits[cursor+1] << 2) |
(bits[cursor+2] << 1) |
bits[cursor+3];
This is going to result in lines and lines of code when reading larger types like int32 and int64 though.
Is it possible to do a cast of some kind and achieve the same result? I'm not concerned about safety at all in this context (these functions will be wrapped into a class that handles that)
Say I wanted to get an uint64_t out of bits, starting at an arbitrary address specified by cursor, when cursor isn't necessarily a multiple of 64; is this possible by a cast? I thought this
uint64_t result = (uint64_t *)(bits + cursor)[0];
Would work, but it doesn't want to compile.
Sorry I know this is a dumb question, I'm quite inexperienced with pointer math. I'm not looking just for a short solution, I'm also looking for a breakdown of the syntax if anyone would be kind enough.
Thanks!
You could try something like this and cast the result to your target data size.
uint64_t bitsToUint64(bool *bits, unsigned int bitCount)
{
uint64_t result = 0;
uint64_t tempBits = 0;
if(bitCount > 0 && bitCount <= 64)
{
for(unsigned int i = 0, j = bitCount - 1; i < bitCount; i++, j--)
{
tempBits = (bits[i])?1:0;
result |= (tempBits << j);
}
}
return result;
}
I want to add a new line in this. This is my sample code:
ui->button->setText(" Tips " + "\n" + TipsCount );
This is the error it shows:
invalid operands of types 'const char [7]' and 'const char [2]' to binary 'operator+'
But when I add to label it gets appended!
ui->label->setText(name + "\n" + City );
Can someone please help me?
This is a very common problem in C++ (in general, not just QT).
Thanks to the magic of operator overloading, name + "\n" gets turned into a method call (couldn't say which one since you don't list the type). In other words, because one of the two things is an object with + overloaded it works.
However when you try to do "abc" + "de", it blows up. The reason is because the compiler attempts to add two arrays together. It doesn't understand that you mean concatenation, and tries to treat it as an arithmetic operation.
To correct this, wrap your string literals in the appropriate string object type (std::string or QString most likely).
Here is a little case study:
QString h = "Hello"; // works
QString w = "World"; // works too, of course
QString a = h + "World"; // works
QString b = "Hello" + w; // also works
QString c = "Hello" + "World"; // does not work
String literals in C++ (text in quotes) are not objects and don't have methods...just like numeric values aren't objects. To make a string start acting "object-like" it has to get wrapped up into an object. QString is one of those wrapping objects, as is the std::string in C++.
Yet the behavior you see in a and b show we're somehow able to add a string literal to an object. That comes from the fact that Qt has defined global operator overloads for both the case where the left operand is a QString with the right a const char*:
http://doc.qt.nokia.com/latest/qstring.html#operator-2b-24
...as well as the other case where the left is a const char* and the right is a QString:
http://doc.qt.nokia.com/latest/qstring.html#operator-2b-27
If those did not exist then you would have had to write:
QString a = h + QString("World");
QString b = QString("Hello") + w;
You could still do that if you want. In that case what you'll cause to run will be the addition overload for both operands as QString:
http://doc.qt.nokia.com/latest/qstring.html#operator-2b-24
But if even that didn't exist, you'd have to call a member function. For instance, append():
http://doc.qt.nokia.com/latest/qstring.html#append
In fact, you might notice that there's no overload for appending an integer to a string. (There's one for a char, however.) So if your TipsCount is an integer, you'll have to find some way of turning it into a QString. The static number() methods are one way.
http://doc.qt.nokia.com/latest/qstring.html#number
So you might find you need:
ui->button->setText(QString(" Tips ") + "\n" + QString::number(TipsCount));