I'm new to C. The string assignment in the following code works:
#include<stdio.h>
int main(void){
char str[] = "string";
printf("%s\n",str);
}
But doesn't work in the following,even I give the index number to the name[]:
#include <stdio.h>
int main(void){
struct student {
char name[10];
int salary;
};
struct student a;
a.name[10] = "Markson";
a.salary = 100;
printf("the name is %s\n",a.name);
return 0;
}
Why does this happen?
You can't assign to an array. Two solutions: either copy the string:
strcpy(a.name, "Markson");
or use a const char pointer instead of an array and then you can simply assign it:
struct {
const char *name;
/* etc. */
};
a.name = "Markson";
Or use a non-const char pointer if you wish to modify the contents of "name" later:
struct {
char *name;
}
a.name = strdup("Markson");
(Don't forget to free the memory allocated by strdup() in this latter case!)
name be a const char* in the second case? :)You cannot do this
a.name[10] = "Markson";
You need to strcpy the string "Markson" to a.name.
strcpy declaration:
char * strcpy ( char * destination, const char * source );
So, you need
strcpy( a.name, "Markson" );
char str[] = "string"; is a declaration, in which you're allowed to give the string an initial value.
name[10] identifies a single char within the string, so you can assign a single char to it, but not a string.
There's no simple assignment for C-style strings outside of the declaration. You need to use strcpy for that.
because in one case you assigning it in a declaration and in the other case you are not. If you want to do the equivalent write:
struct student a = {"Markson", 0};
char name[10];, you get an array with ten elements, numbered 0 to 9 inclusive. Soa.name[10]is outside the bounds of the array. It's not a good idea to try to assign it a value.