Meaning of # symbol as argument in strcmp()

Question

I came across this line of code in legacy code:

#define func(x,y)  if(strcmp(x,#y)==0)

Anyone have an idea of the purpose for the # symbol preceding y?

Answer 1

as mentioned in the comments, this seems like stringification in a c macro.

here is a little example that uses your sample code:

#define doif(x, y) \
    if(strcmp(x,#y)==0) { \
        printf("doing! %s\n",x); \
    }\
    else { \
        printf("not doing!\n"); \
    } 

int main()
{
    char x[] = "test";

    doif (x, test);
    doif (x, something);

    return 0;
}

the stringification operator actually pastes y variable as a string before the compilation stage