C-比较并将char *转换为char

Question

I have the following code:

char* line[255];
....
line[0] = '1'; // filled char by char
....

char charCode = (char)line[i];
if (charCode == '\t' || charCode == '\n' || charCode == ',' || charCode == ' ')
{
      // do someting
}

The IF condition I'm using to find space, comma, tab, or end of line. The warning I get from GCC:

   warning: cast from pointer to integer of different size

Can someone help me spot the issue here? Thanks,

Answer 1

Change:

line[0] = "1"; // filled char by char

to this:

line[0] = '1';

Also, change:

char* line[255];

to:

char line[255];

In C, string literals in double quotes " are of type char * , while literals enclosed in single quotes ' are character literals with type char . Also, you have declared an array of pointers to characters (or an array of strings, essentially) where I think you just wanted an array that can hold 255 characters.

Answer 2

It's too hard to do this in comments. The code should look like this.

char line[255];
....
line[0] = '1'; // filled char by char
....

char charCode = line[i];
if (charCode == '\t' || charCode == '\n' || charCode == ',' || charCode == ' ')
{
      // do someting
}

The array should be char, not char*. That fixes the error. Then the cast is not needed.