Need Explanation for the code. Can someone please explain why the output of the code is -32?

Question

    main() {     
        char c1='a' , c2='A';     
        int i=c2-c1;     
        printf("%d", i); 
    }

What is the output of this code. Please explain why? I know the Answer is -32 but Can someone explain why it's -32?

That's the difference between a lower case letter and its upper case equivalent in the [ASCII table](https://www.asciitable.com/). — 500 - Internal Server Error, Sep 07 '21 at 13:02
I get the answer 64 on my EBCDICBFG2000 computer. So strictly speaking the result isn't guaranteed. Though in practice most people use computers with ASCII or UTF8. — Lundin, Sep 07 '21 at 13:23

John Doe · Answer 1 · 2021-09-07T14:34:45.353

5

Every ascii char is stored as a number between 0 and 127 inclusive, looking at the ascii table you can see that a is stored as 97 while A is stored as 65.

What you are doing is subtracting a from A and then printing it as integer.

Hence why the result is 65-97=-32

edited Sep 07 '21 at 14:34

answered Sep 07 '21 at 13:06

John Doe

1,613
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2

In your answer, you wrote: `"Every char is stored as a number between 0 and 127 inclusive"`. This is not quite correct. On most platforms, the data type `char` is signed and can store values between `-128` to `127`. However, on some platforms, `char` is unsigned and allows values between `0` and `255`. Both variants are permitted by the ISO C standard. ASCII only standardizes the first `127` characters, but the characters `128` to `255` can still be used for other characters. The meaning of these characters normally depends on the code page that is currently in use. – Andreas Wenzel Sep 07 '21 at 13:23
And on yet other platforms, `char` has an even wider range. – John Bollinger Sep 07 '21 at 13:49
@AndreasWenzel you're correct, I wrote the answer quite hastily during my lunch break. I added a small edit to specify `every ascii char` to be more precise – John Doe Sep 07 '21 at 14:35

Vlad from Moscow · Answer 2 · 2021-09-07T13:26:59.063

please explain why the output of the code is -32?

It is because the used compiler uses the ASCII encoding character set where codes of upper case letters are less than codes of corresponding lower case letters by 32.

Here is a demonstrative program.

#include <stdio.h>
#include <ctype.h>

int main(void) 
{
    for ( char upper_case = 'A'; upper_case <= 'Z'; ++upper_case )
    {
        char lower_case = tolower( ( unsigned char )upper_case );
        
        printf( "%c = %d, %c = %d, %c - %c = %d\n",
                upper_case, upper_case, lower_case, lower_case, 
                upper_case, lower_case,
                upper_case - lower_case );
    }
    
    return 0;
}

Its output is

A = 65, a = 97, A - a = -32
B = 66, b = 98, B - b = -32
C = 67, c = 99, C - c = -32
D = 68, d = 100, D - d = -32
E = 69, e = 101, E - e = -32
F = 70, f = 102, F - f = -32
G = 71, g = 103, G - g = -32
H = 72, h = 104, H - h = -32
I = 73, i = 105, I - i = -32
J = 74, j = 106, J - j = -32
K = 75, k = 107, K - k = -32
L = 76, l = 108, L - l = -32
M = 77, m = 109, M - m = -32
N = 78, n = 110, N - n = -32
O = 79, o = 111, O - o = -32
P = 80, p = 112, P - p = -32
Q = 81, q = 113, Q - q = -32
R = 82, r = 114, R - r = -32
S = 83, s = 115, S - s = -32
T = 84, t = 116, T - t = -32
U = 85, u = 117, U - u = -32
V = 86, v = 118, V - v = -32
W = 87, w = 119, W - w = -32
X = 88, x = 120, X - x = -32
Y = 89, y = 121, Y - y = -32
Z = 90, z = 122, Z - z = -32

If your compiler would use the EBCDIC encoding character set then the difference between codes of upper case letters and corresponding lower case letters is equal to 64.

I know the Answer is -32 but

The answer depends on the used encoding character set. It may be for example equal to -32 or -64.:)

Need Explanation for the code. Can someone please explain why the output of the code is -32?

2 Answers2