将 char* 转换为 char* 小写时，不兼容的整数到指针转换从“int”分配给“char *”

Question

I have this annoying warning when I go to compile my code. I'm trying to take a char * and convert it to lower case before I run it through a test. My code compiles just fine but I want to learn how to get rid of this warning. Other posts with this same error don't provide me with what I need to change to remove this error.

//convert to lower case
char *c = "WORD TO LOWER";
char *s[testWordLen];
for (int i = 0; i < testWordLen; i++)
{
    s[i] =  tolower((unsigned char) c[i]);
}

An explanation of what I'm doing wrong would be great too.

Answer 1

You are creating a two-dimensional array using char *s[testWordLen]; ; that is, a pointer to an array of chars. When you dereference it like this: s[i] = tolower((unsigned char) c[i]); , you assign a single character to an array.

The fix is this: declare the variable as char s[testWordLen];