Discussion
What will be the output of the following C code?
#define sqr(x) x*x
main()
{
int a1;
a1=25/sqr(5);
printf("%d",a1);
}
Answer: A
The output of the code shown above is 25. This is because the arithmetic statement takes the form of: 25/5*5 (which is equal to 1). If the macro had been defined as #define sqr(x) (x*x), the output would have been 1.
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