[c] How to Initialize char array from a string

I want to do the following

char a[] = { 'A', 'B', 'C', 'D'};

But I do not want to write these characters separately. I want something like

#define S "ABCD"

char a[] = { S[0], S[1], S[2], S[3] };

But this won't compile (gcc says 'initializer element is not constant').

I tried replacing the #define line with

const char S[] = "ABCD";

But that doesn't seem to help.

How can I do this (or something similar) that lets me write the "ABCD" as a normal 'string', and not as four separate characters?

P.S. It seems that people do not read the question correctly...

I can't get the following code to compile:

const char S[] = "ABCD";
char t[] = { S[0], S[1], S[2], S[3] };
char u[] = { S[3], S[2], S[1], S[0] };

The compilation problem only occurs for me (gcc 4.3, ubuntu 8.10) if the three variables are global. The problem is that C doesn't work like a script languages, so you cannot take for granted that the initialization of u and t occur after the one of s. That's why you get a compilation error. Now, you cannot initialize t and y they way you did it before, that's why you will need a char*. The code that do the work is the following:

#include <stdio.h>
#include <stdlib.h>

#define STR "ABCD"

const char s[] = STR;
char* t;
char* u;

void init(){
    t = malloc(sizeof(STR)-1);
    t[0] = s[0];
    t[1] = s[1];
    t[2] = s[2];
    t[3] = s[3];


    u = malloc(sizeof(STR)-1);
    u[0] = s[3];
    u[1] = s[2];
    u[2] = s[1];
    u[3] = s[0];
}

int main(void) {
    init();
    puts(t);
    puts(u);

    return EXIT_SUCCESS;
}

Perhaps your character array needs to be constant. Since you're initializing your array with characters from a constant string, your array needs to be constant. Try this:

#define S "ABCD"
const char a[] = { S[0], S[1], S[2], S[3] };

That's one of the cases a script to generate the appropriate code might help.

const char S[] = "ABCD";

This should work. i use this notation only and it works perfectly fine for me. I don't know how you are using.

I'm not sure what your problem is, but the following seems to work OK:

#include <stdio.h>

int main()
{
    const char s0[] = "ABCD";
    const char s1[] = { s0[3], s0[2], s0[1], s0[0], 0 };

    puts(s0);
    puts(s1);
    return 0;
}


Microsoft (R) 32-bit C/C++ Optimizing Compiler Version 13.10.3077 for 80x86
Copyright (C) Microsoft Corporation 1984-2002. All rights reserved.
cl /Od /D "WIN32" /D "_CONSOLE" /Gm /EHsc /RTC1 /MLd /W3 /c /ZI /TC
   .\Tmp.c
Tmp.c
Linking...

Build Time 0:02


C:\Tmp>tmp.exe
ABCD
DCBA

C:\Tmp>

Edit 9 June 2009

If you need global access, you might need something ugly like this:

#include <stdio.h>

const char *GetString(int bMunged)
{
    static char s0[5] = "ABCD";
    static char s1[5];

    if (bMunged) {
        if (!s1[0])  {
            s1[0] = s0[3]; 
            s1[1] = s0[2];
            s1[2] = s0[1];
            s1[3] = s0[0];
            s1[4] = 0;
        }
        return s1;
    } else {
        return s0;
    }
}

#define S0 GetString(0)
#define S1 GetString(1)

int main()
{
    puts(S0);
    puts(S1);
    return 0;
}

This compiles fine on gcc version 4.3.3 (Ubuntu 4.3.3-5ubuntu4).

const char s[] = "cheese";

int main()
{
    return 0;
}

Here is obscure solution: define macro function:

#define Z(x) \
        (x==0 ? 'A' : \
        (x==1 ? 'B' : \
        (x==2 ? 'C' : '\0')))

char x[] = { Z(0), Z(1), Z(2) };

Weird error.

Can you test this?

const char* const S = "ABCD";
char t[] = { S[0], S[1], S[2], S[3] };
char u[] = { S[3], S[2], S[1], S[0] };

Simply

const char S[] = "ABCD";

should work.

What's your compiler?

Another option is to use sprintf.

For example,

char buffer[50];
sprintf( buffer, "My String" );

Good luck.