how string (command line) are stored in char**argv and int *argv?

Question

First snippet:

#include<stdio.h>
int main(int argc, char **argv)
{
int i;
for(i=1; i<argc; i++)
    printf("%s\n", argv[i]);
return 0;
}

load time input :

./a.out devang samir

output:

devang
samir

Second snippet:

#include<stdio.h>
int main(int argc, int *argv)
{
int i;
for(i=1; i<argc; i++)
    printf("%s\n", argv[i]);
return 0;
}

load time input :

./a.out devang samir

output:

devang
samir

in both case, i got output same, but why?

• in first case how the strings (command line ) are stored in char** argv ?
• in second case how the string (command line ) are stored in int * argv...?

Answer 1

The C11 standard specifies the function signature for main() in chapter §5.1.2.2.1 as

The function called at program startup is named main. The implementation declares no prototype for this function. It shall be defined with a return type of int and with no parameters:

int main(void) { /* ... */ }

or with two parameters (referred to here as argc and argv, though any names may be used, as they are local to the function in which they are declared):

int main(int argc, char *argv[]) { /* ... */ }

or equivalent;[...]

and regarding the constrains,

If the value of argc is greater than zero, the array members argv[0] through argv[argc-1] inclusive shall contain pointers to strings,[...]

Then, in your second case,

 int main(int argc, int *argv)

char* and int ( for argv[n], in general) are being different types altogether (i.e, not compatible type), your second program invokes undefined behavior.

To elaborate, in case of the functions without having a prototype, the parameters passed to the function while calling should exactly match the type of expected arguments.

Quoting the standard, chapter §6.5.2.2

[...] If the function is defined with a type that does not include a prototype, and the types of the arguments after promotion are not compatible with those of the parameters after promotion, the behavior is undefined.