Is casting implicit in C when computing a sum into a larger value?

Question

I am working on a 16-bit processor, so most of my data is in 16-bit except where necessary.

If I have two 16-bit variables a and b and sum them together into a 32-bit variable, what will the compiler do?

uint16_t x, y;
uint32_t z;
x = 65504;
y = 65503;
z = x + y;

Will the result in z be identical to z = (uint32_t)x + (uint32_t)y, or do I need to cast the result?

I have tried this on my compiler and the casts don't seem to make any difference, but this might just be a compiler oddity for this little embedded processor.

Answer 1

In C99 onwards,^* operands to arithmetic operands are implicitly promoted to be at least as big as int, as part of the usual arithmetic conversions. So the behaviour of your code depends on the native size of int on your platform.

If your int is 32-bit, then your code is equivalent to:

z = (int)x + (int)y;

If your int is 16-bit, then no conversion will occur, and you would get incorrect results due to integer overflow.

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_{* Prior to C99, the promotion rules were less well-defined (although I forget the details).}