Thread: Help with cast in C!

If you use a plain cast, then the lowest 8 bits are being used. This is probably the answer you wanted.

Slightly more complex situation:
If you cast a pointer to another pointer type, and then dereference it, then it's the 8 bits with the lowest address you get. Depending on the byte-order of the machine, you get EITHER the lowest or the highest.

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Mats

Originally Posted by Tirania

I wanna cast 32bit integer into 8bit char value, and the problem is that I don't know which 8bits of the integer would be taken and stored in the 8bit char variable?! The most significant or least significant?!!

According to the C Standard, "if the destination type is signed, the value is unchanged if it can be represented in the destination type (and bit-field width); otherwise, the value is implementation-defined."

Frankly, it would be clearer to use bitwise operations to extract those bits that you want.

Originally Posted by laserlight

According to the C Standard, "if the destination type is signed, the value is unchanged if it can be represented in the destination type (and bit-field width); otherwise, the value is implementation-defined."

Frankly, it would be clearer to use bitwise operations to extract those bits that you want.

Ah, good point - it is undefined, so the compiler is allowed to do whatever it likes with it. However, I have not yet "met" a compiler that doesn't do what I explained above.

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Mats

Originally Posted by matsp

Ah, good point - it is undefined, so the compiler is allowed to do whatever it likes with it. However, I have not yet "met" a compiler that doesn't do what I explained above.

I had in mind an iterative bitwise copy of the 8 bits in question, but come to think of it, the implementation defined behaviour can be avoided with just one bitwise and, e.g.,

Code:

char c = (char)(i & 0xff); // where i is the source integer