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I thought the following code might cause an overflow since a * 65535 is larger than what unsigned short int can hold, but the result seems to be correct.

Is there some built-in mechanism in C that stores intermediary arithmetic results in a larger data type? Or is it working as an accident?

unsigned char a = 30;
unsigned short int b = a * 65535 /100;
printf("%hu", b);
phuclv
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user12072031
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  • what bit width is `short int` on your compiler? compiler usually use its native bitwidth for constant expresion evaluations so its most likely computed on 32 or 64 bits and the result is then truncated to target bitwidth... the same applies to float/double unless you explicitly cats type (`1ull`, ... `1.0, 1.0f`,... ) – Spektre Apr 28 '21 at 07:24

1 Answers1

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It works because all types narrower than int will go under default promotion. Since unsigned char and unsigned short are both smaller than int on your platform, they'll be promoted to int and the result won't overflow if int contains CHAR_BIT + 17 bits or more (which is the result of a * 65535 plus sign) 22 bits or more (which is the number of bits in 30 * 65535 plus sign). However if int has fewer bits then overflow will occur and undefined behavior happens. It won't work if sizeof(unsigned short) == sizeof(int) either

Default promotion allows operations to be done faster (because most CPUs work best with values in its native size) and also prevents some naive overflow from happening. See

phuclv
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