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;
}
Add a comment
|
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";
Comments
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);
Comments
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;
}
