Set string variable in C from argv[1]

I am trying to learn C, and I wonder why this doesn't work?

#include <stdio.h>
int main(int argc, char *argv[])
{ char testvar[] = argv[0]; //do something with testvar return 0;
}

3 Answers

You could do this instead:

char *testvar = argv[0];

Or maybe:

char *testvar = strdup(argv[0]);
/* Remember to free later. */

As pmg notes, strdup isn't standard. Implementing it using malloc + memcpy is a nice exercise.

Or even:

char testvar[LENGTH];
if (strlen(argv[0]) >= LENGTH) fprintf(stderr, "%s is too long!\n");
else strcpy(testvar, argv[0]);

But then again you might be looking for:

char testvar[] = "testvar";

That syntax is valid only to initialize a char array from a literal, i.e. when you explicitly write in the source code what characters must be put in the array.

If you just want a pointer to it (i.e. "another name" to refer to it) you can do:

char * testvar = argv[0];

if instead you want a copy of it you have to do:

size_t len = strlen(argv[0]);
char * testvar = malloc(len+1);
if(testvar==NULL)
{ /* allocation failed */
}
strcpy(testvar, argv[0]);
/* ... */
free(testvar);

You need a constant to initialize an object. argv[0] is not a constant.
"foobar" below is a constant

#include <stdlib.h>
#include <string.h>
int main(int argc char **argv) { char array[] = "foobar"; /* array has 7 elements */ char *testvar; testvar = malloc(strlen(argv[0]) + 1); /* remember space for the NUL terminator */ if (testvar != NULL) { strcpy(testvar, argv[0]); /* do something with testvar */ free(testvar); } else { /* report allocation error */ } return 0;
}