CODEWITHSUNDEEP
cintegerinteger-overflow
Abhijeet Gautam
Abhijeet Gautam

Reputation: 9

What determines the data type of a variable- the declarative keyword (short int) or the format specifier (%hd)?

I have been trying to understand the integer overflow in C-programming. I am confused about whether the final value output depends on the initial datatype given to the variable during declaration or the format specifier used to print the variable.

`

int main() {
    short int a = 32771;
    printf("%d\n", a); // O/P : -32765
    printf("%u\n\n", a); // O/P : 4294934531

    int b = 32771;
    printf("%hd\n", b); // O/P : -32765
    printf("%hu", b); // o/p : 32771

    return 0;
}

`

a is declared a short-integer at the very start, but initialized with a value that overflows a short-integer range. The printf("%d\n", a) statement gives output considering a as a signed short-integer (2 byte or 16 bits), whereas the printf("%u\n\n", a) statement gives output considering it as an unsigned integer (4 bytes or 32 bits)

b is declared a integer (4 bytes or 32 bits) at the very start, and initialized with a value well within the integer range. The printf("%hd\n", b) statement gives output considering b as a signed short-integer (2 byte or 16 bits), whereas the printf("%hu", b) statement gives output considering it as an unsigned short-integer.

Please explain this discrepancy. What exactly determines the final output value?

Upvotes: 0

Views: 131

Answers (2)

Eric Postpischil
Eric Postpischil

Reputation: 222302

short int a = 32771;

In an initialization, the initial value, 32,771, is converted to the type of the object being initialized, short int, per C 2018 6.7.9 11 (paragraph 11 of clause 6.7.9 of the 2018 C standard) and 6.5.16 2.

In your C implementation, short int is 16 bits and can represent values from −32,768 to +32,767. So it cannot represent 32,771. When a value of integer type is converted to a signed integer type and cannot be represented in the new type, the result is implementation-defined or an implementation-defined signal is raised, per 6.3.1.3 3. This means the C implementation is required to document what it does for this. A common behavior is to wrap the result modulo 216, so the result is 32,771 − 65,536 = −32,765.

printf("%d\n", a);

printf is declared with ... for arguments after its first. In such a call, the default argument promotions are performed on the trailing arguments, per 6.5.2.2 6. These promotions promote a short int to an int. The %d in the format string tells printf to expect an int, so it gets that int, −32,765, and prints it.

printf("%u\n\n", a);

Again a is promoted to int. However, %u tells printf to expect an unsigned int. When the types do not match, the behavior is not defined by the C standard, per 7.21.6.3 2 and 7.21.6.1 9. However, it is common that printf will reinterpret the bits of an int as a representation of an unsigned int, which is 32 bits in your C implementation. When a C implementation uses two’s complement, the bits that represent −32,765 in a 32-bit int are FFFF800316. Interpreted as an unsigned int, these bits represent 4,294,934,531, so that is what is printed.

printf("%hd\n", b);

%hd tells printf to expect an int but to convert its value to short int before printing, per 7.21.6.1 7. So the conversion of 32,7761 to short int produces −32,765, and that is what is printed.

printf("%hu", b);

%hu tells printf to expect an int but to convert its value to unsigned short int before printing, per 7.21.6.1 7. (This can vary in other C implementations; printf will expect whatever type an unsigned short int is promoted to by the integer promotions in 6.3.1.1 2, which is int in your C implementation but could be unsigned int.) Since 32,771 can be represented in unsigned short int, the conversion does not change the value, and that is what is printed.

Upvotes: 0

RTFM
RTFM

Reputation: 1

The type of the variable is determines when you create it.

What append is when printf() pars the const char * format he find %u in the string.

That say to him, he need to print a unsigned int, and he gonna read it with a va_arg() call. (see man)

And he print the unsigned int he create, with a write() call.

Just for info :

An overflow condition may give results leading to unintended behavior. In particular, if the possibility has not been anticipated, overflow can compromise a program's reliability and security.

Wikipedia

If you want to dig more about bits and overflow you can print bits :

#include <stdio.h>
#include <stddef.h>
#include <stdint.h>

void print_bits(short int x)
{
        printf("%d : \n", x);
        for (int16_t i = sizeof(x) * 8 - 1 ; i >= 0; i--)
        {
                printf("%d", (x >> i) & 1);
                if (i % 8 == 0)
                        printf(" ");
        }
        printf("\n");
}

int main(void)
{
        short int a = 32767;
        print_bits(a);
        a++;
        print_bits(a);
        return (0);
}

That give us something like that :

➜ RTFM ./a.out
32767 :
01111111 11111111
-32768 :
10000000 00000000

Upvotes: -4

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