Difference between char *argv[] and char **argv for the second argument to main()

c arrays pointers parameter-passing argv
Jhansi Rani picture Jhansi Rani · Nov 30, 2014 · Viewed 51k times · Source

CODE 1

#include<stdio.h>

int main(int argc, char *argv[])
{
int j;
printf("%d", argv[1][0]);
return 0;
}

CODE 2

#include<stdio.h>

int main(int argc, char **argv)
{
int j;
printf("%d", argv[1][0]);
return 0;
}

CODE 1 and CODE 2 both give same output. but argument 2 of main function in CODE 1 and CODE 2 are different. Array of pointers are created above data section at compile time. argv is array of pointers. Then we should declare argument in main function as pointer to pointer to character i.e., **argv. How it is correct to declare as in CODE 1?

Answer

bmargulies picture bmargulies · Nov 30, 2014

It is fundamental to c that char** x and char* x[] are two ways of expressing the same thing. Both declare that the parameter receives a pointer to an array of pointers. Recall that you can always write:

 char *parray[100];
 char **x;

 x = &parray[0];

and then use x identically.

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