CODEWITHSUNDEEP
c++cmacrosstrcmp
user1628622
user1628622

Reputation:

Meaning of # symbol as argument in strcmp()

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?

Upvotes: 0

Views: 129

Answers (2)

Pranit Kothari
Pranit Kothari

Reputation: 9839

First of all you posted wrong or in complete code. #y should be used with macro definition, not while using macro. 

#define MAC(STR) #STR    

int main(int argc, char* argv[])
{
    printf(MAC(ME));//prints ME
    printf(MAC("ME"));//prints "ME"

    return 0;
}

Here I have defined MAC macro which takes one argument. I did it's stringification.

Also see second printf, it exactly prints string. So you need not to give pair of "".

Upvotes: 1

Shlomi
Shlomi

Reputation: 4756

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

Upvotes: 3

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