CODEWITHSUNDEEP
cmacrosc-preprocessor
progbrain
progbrain

Reputation: 33

understanding a C Preprocessor Macro's output vs a line code

#define AD(x,y) (x+y)
int main()
{
    int x1=5,y1=2,z1;
    int x2=5,y2=2,z2;
    z1 = AD(x1,++y1);
    z2 = (x2+++y2) ;
    printf("%d %d %d\n",x1,y1,z1);
    printf("%d %d %d\n",x2,y2,z2);
}

why the output is different? the first case is : 5 3 8 and the second is : 6 2 7

Upvotes: 3

Views: 72

Answers (1)

Vlad from Moscow
Vlad from Moscow

Reputation: 310980

This expression

z2=x2+++y2;

is parsed by the compiler like

z2 = x2++ + y2;

From the C Standard (6.4 Lexical elements)

4 If the input stream has been parsed into preprocessing tokens up to a given character, the next preprocessing token is the longest sequence of characters that could constitute a preprocessing token.

So these tokens +++ are parsed like ++ and +. The expression with the macro

z1=AD(x1,++y1);

is parsed by the compiler like

z1 = x1 + ++y1;

The compiler already formed these sets of tokens x1 and ++y1 due to the comma between the tokens.

So these two statements are different.

Upvotes: 5

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