Question

I have asked a similar question about command line arguments in C++ some hours ago. Now I have another problem, as far as I know, command line arguments will be saved like a string in argv array. so comparing theme with a string should be logical, but it does not work in the way I want, look at this code:

#include <iostream>
using namespace std;

int main(int argc,char** argv)
{
    if (argv[2]=="stack") cout << "right";
    cout << argv[2];
    return 0;
}

now I pass this command to my compiled application named zero.exe;

zero.exe stack

the output should be “rightstack”, but if command will skip and only cout << argv[2]; will execute, so only stack will be shown on monitor. it shows "stack" is saved into argv[2], so if (argv[2]=="stack") should work, but it is not. where is the problem?

Answer 1

For historical reasons, the arguments are passed as C-style strings; that is, each is a pointer to an array of characters, with a zero-valued character to mark the end. Similarly, a string literal (like "stack") is a simple array of characters.

Your code compares two pointers, which will not be equal even if the string values are equal. To compare the strings, either convert one (or both) to std::string:

#include <string>

std::string arg2(argv[2]);
if (arg2=="stack") std::cout << "right\n";

or use the C library function for comparing C-style strings; this might be more efficient, but is also harder to read:

#include <cstring>

if (std::strcmp(argv[2], "stack") == 0) std::cout << "right\n";

Also, the arguments are counted from 1, with the program name as argv[0], so you probably want to be testing argv[1] rather than argv[2].