In C, why can't I assign a string to a char array after it's declared?

问题

This has been bugging me for a while.

struct person {
       char name[15];
       int age;
};
struct person me;
me.name = "nikol";

when I compile I get this error:

error: incompatible types when assigning to type ‘char[15]’ from type ‘char *’

am I missing something obvious here?

回答1:

Arrays are second-class citizens in C, they do not support assignment.

char x[] = "This is initialization, not assignment, thus ok.";

This does not work:

x = "Compilation-error here, tried to assign to an array.";

Use library-functions or manually copy every element for itself:

#include <string.h>
strcpy(x, "The library-solution to string-assignment.");

回答2:

me.name = "nikol"; is wrong !! you need to use strcpy()

when you do x = "Some String", actually you are putting the starting address of the static string "Some String" into variable x. In your case, name is a static array, and you cannot change the address. What you need, is to copy your string to the already allocated array name. For that, use strcpy().

回答3:

First of all, you need to know the following points:

In C, text strings are just arrays.
In C, array variables are basically just pointers.

So, char mytext[12]; is essentially just declaring a char pointer called mytext that stores the address of the first (zero'th) element of the array/string.

This code is therefore valid:

#include <stdio.h>
int main(int argc, char *argv[])
{
    const char a[] = "Hello";
    const char *b = a;
    printf("%s\n", b);
    return 0;
}

The important thing to note here is that re-assigning b doesn't change the contents of whatever it points to - it changes the thing that it points to.

However, there are cases where arrays and pointers behave differently. In the example above, a cannot be reassigned. If you try, you'll get an error.

To go back to you original example, this structure:

struct person{
    char name[15];
    int age;
};

...can be thought of as a 19-byte structure* of which the first 15 bytes are earmarked for storing a string. The name attribute stores the address of the first byte, so you know where those 15 bytes live in memory - you just need to write something useful into them.

This is where functions such as sprintf() or strcpy() come into play - they copy data into the address defined by name rather than re-assigning name itself.

* Assuming that sizeof(int) is 4 and the structure is not padded, of course...

回答4:

The initialization:

char name[15]

is creating a char array, 15 bytes in size.

"nikol" is a string literal, which is type 'char *', i.e. a pointer to a string. They are incompatible types. You need to copy the string into the array, like so:

strcpy(me.name, "nikol")

回答5:

Use strcpy() function (of string.h library) :)

main(){
struct person{
       char name[15];
       int age;
};
struct person me;
strcpy(me.name,"nikol");
}

回答6:

You are trying to change the memory address of an array pointer (by giving me.name = "nikol") which is impossible. Name of array (me.name) is a fixed pointer and cannot be changed to point to a new location of string ("nikol") because memory address of an array pointer is on the first member of that array. If you want to assign a string after declared (me.name = "nikol"), use external pointer *name instead because external pointer floats outside array so its value can be changed on-the-fly.

struct person {
       char *name;
       int age;
};
struct person me;
me.name = "nikol"; // no error

来源：https://stackoverflow.com/questions/26693537/in-c-why-cant-i-assign-a-string-to-a-char-array-after-its-declared

标签

arrays

string

initialization

assign