Unexpected output of Program

[Ayan Dasgupta]

int a=5;
int b=a++ + ++a + a++;
printf("%d",b);
 

the output for this code is 19

int a=5,b;
b=a++ + ++a + a++;
printf("%d",b);

but the output for this code is 18.

why?

[weaknessforcats]

The C language standard provides no rules for when a compiler can change variable values inside a single statement. As a result, if you change the value of a variable more than once in a single statement, the compiler can make the changes in any order it chooses.

This is termed indeterminate results.

[AjayGohil]

Hi,

In my system both program give same answer 19,so may be it is depend on compiler.You use a for multiple time so,it is depend on compiler.There is no specific rules for compiler.

[lewish95]

I recommend this method over a for (t = 0; t < b; t++). Imagine if you were using 64-bit math, the for loop would take a long time, even on a computer. OP doubling method of a=a+a was on track, but incomplete.

unsigned a;
unsigned b;
unsigned product = 0;
scanf("%u%u", &a, &b);
while (b > 0) {
if (b & 1) {
unsigned OldProduct = product;
product += a;
if (product < OldProduct) {
printf("\nOverflow\n");
break;
}
b >>= 1; // Halve b
a += a; // Double a - avoid using multiplication
}
printf("\nProduct = %u\n", product);