4字节char数组为整数，将在C的case语句中使用

Question

How can I use a char array with length 4 which is a string literal) to represent a 4 byte integer?

I want to use a string literal with length 4 to substitute a 4 byte integer, which is to be used in this case statement. I want to write like:

case "\x0ATst":
    printf("FOUND TIME STAMP");
    break;

not:

case 0x7473540A:
    printf("FOUND TIME STAMP");
    break;

As the primary code is more human readable.

I searched for "string to int" on google, but only answers like atoi() were found.

Answer 1

switch only works with integer constants, so it isn't very flexible at all. You can't use case with expressions, strings nor with any other type than int .

But in this case it doesn't matter, because you should have neither form, because neither of them is very readable. Instead use case STAMP: , where STAMP is some suitable, self-explaining name. It could be a #define , const or enum .

As for how to convert from a 4 character string to an int , there's various reasons why you can't simply do *(int*)"\\x0ATst" (alignment, strict aliasing). There's a work-around for that, by using a union:

typedef union
{
  char str[4];
  unsigned int val;
} conv_t;

...

(conv_t){.str = "\x0ATst" }.val

Please note that this code depends heavily on CPU endianess, so it is still not very good practice.