Is typecast required in malloc? [duplicate]

问题:

This question already has an answer here:

• Do I cast the result of malloc? 26 answers

What is the use of typecast in malloc? If I don't write the typecast in malloc then what will it return? (Why is typecasting required in malloc?)

回答1:

I assume you mean something like this:

int *iptr = (int*)malloc(/* something */);

And in C, you do not have to (and should not) cast the return pointer from malloc. It's a void * and in C, it is implicitly converted to another pointer type.

int *iptr = malloc(/* something */);

Is the preferred form.

This does not apply to C++, which does not share the same void * implicit cast behavior.

回答2:

You should never cast the return value of malloc(), in C. Doing so is:

• Unnecessary, since void * is compatible with any other pointer type (except function pointers, but that doesn't apply here).
• Potentially dangerous, since it can hide an error (missing declaration of the function).
• Cluttering, casts are long and often hard to read, so it just makes the code uglier.

So: there are no benefits, at least three drawbacks, and thus it should be avoided.

回答3:

You're not required to cast the return value of malloc. This is discussed further in the C FAQ: http://c-faq.com/malloc/cast.html and http://c-faq.com/malloc/mallocnocast.html .

回答4:

Just because malloc returns a void* and since void* has not defined size you can't apply pointer aritmetic on it. So you generally cast the pointer to the data type your allocated memory block actually points.

回答5:

The answers are correct, I just have an advice:

• don't play with pointers - which malloc() returns - too much, cast them to a specified type asap;
• if you need to do some math with them, cast them to char*, so ptr++ will mean what you except: add 1 to it (the size of the datatype will be added, which is 1 for char).