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some error in output in using macro in C

my code is:-

#include<stdio.h>
#include<conio.h>

#define sq(x) x*x*x

  void main()
  {
    printf("Cube is : %d.",sq(6+5));
    getch();
  }

The output is:-

Cube is : 71.

now please help me out that why the output is 71 and not 1331...

thank you in advance.

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1 Answer

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by (71.8m points)

Always shield your macro arguments with parenthesis:

#define sq(x) ((x) * (x) * (x))

Consider the evaluation without the parenthesis:

6 + 5 * 6 + 5 * 6 + 5

And recall that * has a higher precedence than +, so this is:

6 + 30 + 30 + 5 = 71;

Get to know the precedence rules if you don't already: http://en.cppreference.com/w/cpp/language/operator_precedence


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