What does *= do?

Question

Hey I am kinda new to C and I wanted to ask why this prints out 4 instead of 260?

#include <stdio.h>

int main()
{
    unsigned char x = 130;
    x *= 2;
    printf("%d\n", x);
}

Answer 1

The *= operator is called multiplication assignment operator and is shorthand for multiplying the operand to the left with the operand to the right and assigning the result to the operand to the left. In this case, it's the same as:

x = x * 2;

Here integer promotion first takes place and the result of x * 2 is indeed 260.

However, an unsigned char can usually only carry values between 0 and 255 (inclusive) so the result overflows (wraps around) when you try assigning values above 255 and 260 % 256 == 4.

Answer 2

The unsigned char can contain only from 0 to 256. When the value is over 256, the value is forced from 0 to 256.

Answer 3

According to language specification unsigned char data type takes 1 byte of memory so it can store values 0-255 and if you a value greater than 255 it will overflow beginning from zero. So 260(130 * 2) - 256 = 4 is assigned to your variable. https://learn.microsoft.com/en-us/cpp/cpp/cpp-type-system-modern-cpp?view=msvc-170#fundamental-built-in-types

Answer 4

In the statement with the compound assignment operator

x*=2;

that is equivalent to

x = x * 2;

the operand x of the expression is converted to the type int due to the integer promotions and the result pf the expression is indeed equal to 260. But the result is assigned to the variable x that has the type unsigned char.

unsigned char x=130;

The value 260 can not be stored in such an object. As 260 is internally as an integer is represented like

0x00000104

then only the last byte having the value 0x4 is stored in the object and this value is outputted.

You could get the expected result if for example the type of the variable x would be changed at least like

unsigned short x=130;

Answer 5

x *= 2 uses the multiplication assignment operator, a short hand notation for x = x * 2 where x is only evaluated once.

The behavior of this small program is non trivial:

• The multiplication is performed after integer promotion of the arguments, which means the value of x, 130 is promoted to type int and the multiplication gives 260 as an int.
• This value is converted to the destination type, unsigned char by repeatedly subtracting UCHAR_MAX+1, which is 256 on your system, until reaching a value in the range of the destination type. Hence x becomes 260 % 256 = 4.
• Because it has an integer type smaller than int, x is promoted to int when passed to printf, so the format %d, which expects an int value, has defined behavior and produces 4.

Note that for exotic architectures (eg: digital signal processing chips) where unsigned char has more than 8 bits, some of the above discussion is irrelevant and the printf may print 260 or even have undefined behavior (if sizeof(int) == 1).