What is the %d value of char?

Question

Can someone explain the result of this? Please explain step by step clearly. Thanks a lot.

Why it's -16, -6, - 22? How compiler got these value?

Here's the code:

int main(void)
{
    char i=240, j=250, sum;
    sum = i+j;
    printf("Result is: %d + %d = %d\n",i,j,sum);

    return 0;
}

Here is the result: Result is: -16 + -6 = -22

Char is 8 bits. Take the result % 256, check if the high bit is set. — Mad Physicist, Jun 15 '19 at 07:18

Mohammad Ansari · Answer 1 · 2019-06-15T07:32:23.900

1

Because Char is 8 bits and it's capacity is 255

When you add i+j you put ‭490‬ in sum

Then sum variable overflow and it generate negative number

Use int instead of char for sum variable it will solve your problem

edited Jun 15 '19 at 07:32

answered Jun 15 '19 at 07:20

Mohammad Ansari

Thanks. But how compiler got these value? any formula? – 田睿霖 Jun 15 '19 at 07:26
@田睿霖 Just count if your number is 254 then your char variable will be 254-256=-2 or if your number is 253 then char variable will be 253-256 = -3 – Mohammad Ansari Jun 15 '19 at 07:29
@田睿霖 i add link to my answer you can read it and you will understand why – Mohammad Ansari Jun 15 '19 at 07:32
"sum variable overflow " --> No. -16 + -6 --> -22. No overflow in the addition nor assignment to `sum`. – chux - Reinstate Monica Jun 15 '19 at 09:00

score 0 · Accepted Answer · answered Jun 15 '19 at 07:18

0

char i=240

Overflow. char = 8 bits. When signed, it ranges from -128 to 127. So this converts it to negative.

Suggestion: read your compiler's warnings carefully.

answered Jun 15 '19 at 07:18

How the compiler got the -16? thx. – 田睿霖 Jun 15 '19 at 07:23
unsigned 255 = signed -1. unsigned 254 = signed -2, etc. – Michael Chourdakis Jun 15 '19 at 07:23
1

There is no overflow, just implementation-specified conversion – Antti Haapala -- Слава Україні Jun 15 '19 at 07:33
That answer would confuse him. Sometimes it's best to oversimplify things. – Michael Chourdakis Jun 15 '19 at 07:35

2 Answers2