I was trying to declare an int array in C++ and found this problem. The following code runs fine on g++ compiler but the compilation fails on Visual Studio. I was following Bruce Eckel and found this code.
#include<iostream>
int main()
{
const int j = std::cin.get();
char buf[j];
}
Keeping j just an int would be a problem, that I understand. Since the value of j would be const during the run-time, the program should get compiled. Please correct me if I am wrong anywhere.
Since the value of
jwould beconstduring the run-time, the program should get compiled.
No, the const-ness of j is irrelevant here. C++ currently only supports statically-sized C-arrays. Its size must be a compile-time constant.
If you want an array of dynamic size, use std::vector.
The fact that g++ by default compiles this is a bit unfortunate (for compatibility). You should use the -pedantic flag when using g++ to ensure that such compiler extensions aren’t enabled (using compiler extensions of course isn’t bad in itself, but in this case there’s not really any advantage).
-pedantic-errors or -pedantic -Werror to disable the extensions. Use -pedantic to get warnings about use of extensions.You are trying to define buf as a variable-length array. This is a feature of C (not C++) that is supported by g++ as a non-standard extension. Evidently your other compiler does not support it.
I would suggest turning buf into std::vector<char> (or indeed std::string?)
Variable length arrays are C99 feature both gcc and clang supports them as an extension in C++ but Visual Studio never did and even though they recently added support for C99 is it not supported in C++
Since you are developing in C++ unless you have a good reason to not use it then std::vector or std::string should be sufficient.
I had the same problem. It seems very difficult to create an array of which its length is stored as a variable. What I did is create an additional function:
void Class:: initMyArray(const int size) {
myArray = new int[size]
}
Now, you just have to call that function and give it your variable. I'm not sure whether this is a proper solution though (I'm not a C++ expert).
jis not a constant expression if you had usedconst int j = 10it would have worked since using a literal would have made it a constant integral expression.