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Why the output of b is 16 not 18

#include <stdio.h> int main() { int a[5],i,b=16; for(i=0;i<5;i++) a[i]=2*i; f(a,b); for(i=0;i<5;i++) printf("\n%d",a[i]); printf("\nb=%d",b); return 0; } f(int *x,int y) { int i; for(i=0;i<5;i++) *(x+i)+=2; y+=2; printf("\ny=%d",y); }

16th Apr 2018, 10:56 AM
Nitesh Yadav
Nitesh Yadav - avatar
1 Answer
0
You initialize b to 16 and never change its value. Remember the difference between int *x and int x when using them as arguments for functions. int x does not change that variable, in this case, b. Does this make sense?
16th Apr 2018, 11:02 AM
J.G.
J.G. - avatar