Error in macro expansion

Question

I have been trying to understand macro expansion and found out that the second printf gives out an error. I am expecting the second print statement to generate the same output as the first one. I know there are functions to do string concatenation. I am finding it difficult to understand why first print statement works and the second doesn't.

#define CAT(str1, str2) str1 str2

void main()
{
    char *string_1 = "s1", *string_2 = "s2";
    printf(CAT("s1", "s2"));
    printf(CAT(string_1, string_2));
}

[Possible duplicate](http://stackoverflow.com/questions/10295634/c-string-concatenation-of-constants). — Lundin, Nov 02 '15 at 14:26

score 8 · Answer 1 · edited Nov 02 '15 at 12:42

8

Concatenating string literals, like "s1" "s2", is part of the language specification. Just placing two variables next to each other, like string_1 string_2 is not part of the language.

If you want to concatenate two string variables, consider using strcat instead, but remember to allocate enough space for the destination string.

edited Nov 02 '15 at 12:42

Am_I_Helpful

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answered Nov 02 '15 at 12:39

Some programmer dude

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score 1 · Accepted Answer · answered Nov 02 '15 at 12:40

1

Try to do the preprocessing "by hand":

CAT is supposed to take 2 input variables, and print them one after the other, with a space between. So... if we preprocess your code, it becomes:

void main()
{
    char *string_1 = "s1", *string_2 = "s2";
    printf("s1" "s2");
    printf(string_1 string_2);
}

While "s1" "s2" is automatically concatenated to "s1s2" by the compiler, string_1 string_2 is invalid syntax.

answered Nov 02 '15 at 12:40

Amit

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I thought it would be the same as the first statement as the value in string_1 is "s1" and string_2 is "s2" – Ajit Nov 02 '15 at 12:44
All the preprocessor does is manipulate the source code itself, once that's done, the code goes through normal compilation, and must be valid for that purpose. As I explained, `string_1 string_2` is invalid, and can't compile. – Amit Nov 02 '15 at 12:46
@Pete `string_1` is a variable to be precise , and `"s1"` is a string literal . Both are treated in different manner . – ameyCU Nov 02 '15 at 12:46
@Pete `string_1` doen's have the value "s1". It's a pointer that points to a zero terminated string "s1". – Jabberwocky Nov 02 '15 at 12:47

Error in macro expansion

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