How can I replace characters in a string using a pointer? (in c code)

Question

Here's my code:

#include <stdio.h>
#include <string.h>

unsigned char code[] = "Hello world!\n";
main()
{

        printf("String Length:  %d\n", strlen(code));
        printf("Original String: %s\n", code);

        char &code[7] = "W";    
        char &code[8] = "a";
        char &code[9] = "l";
        char &code[10] = "e";
        char &code[11] = "s";

        printf("New String: %s\n", code);

}

To start with, you do know that array indexes in C starts at zero? So the index of the 'W' in "Hello World" is 6. Although you would notice that very quickly with this program after fixing the syntax. — Some programmer dude
– Some programmer dude, Commented Aug 26, 2015 at 14:34
Also, arrays and pointers are very often interchangeable, and you can use array indexing syntax to access elements when using pointer, as well as using an array as a pointer. — Some programmer dude
– Some programmer dude, Commented Aug 26, 2015 at 14:37

Andy Thomas · Accepted Answer · 2015-08-26 14:41:56Z

2

You can specify a zero-based array index:

   code[6] = 'W';    
   code[7] = 'a';
   code[8] = 'l';
   code[9] = 'e';
   code[10] = 's';

Character literals are specified with single quotes rather than double.

The array variable is a synonym for the address of the first element. If you specifically want to use pointer syntax, you can replace code[i] with *(code + i). For example:

   *(code + 6) = 'W';    
   *(code + 7) = 'a';
   *(code + 8) = 'l';
   *(code + 9) = 'e';
   *(code + 10) = 's';

edited Aug 26, 2015 at 14:41

answered Aug 26, 2015 at 14:34

Andy Thomas

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