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In the program of any pointer variable we often use : 

float *x;
x=(float*)malloc(a*sizeof(long int));

I want to know why we use (float*) in front of malloc?

Marco Leogrande
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user2100721
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    See [Should I explicitly cast malloc()'s return value?](http://stackoverflow.com/q/953112/1168156) – LihO Feb 22 '13 at 20:03
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    You use it because you haven't read [this article](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc). –  Feb 22 '13 at 20:03
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    Why do you use `sizeof(long int)` for allocating an array of `float` ?? – Martin R Feb 22 '13 at 20:06

2 Answers2

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Malloc returns a pointer to void.

(float*) casts from a pointer to void to a pointer to float

In C this is not necessary, in C++ it is, so some people recommend that to make your code compatible with C++ compilers.

But you don't need to do that. (and some C fans are against it)

speeder
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malloc will give you a pointer to void that you can't use for anything related to things you want to do with a float. To be able to use the variable allocated at the returned memory location, you need to cast it to a float* so you can dereference that pointer and use it as a float.

But, as you've written your question, you should cast the return value of malloc to float* and then immediately dereference it before assigning it to x, since you've not declared x as a pointer to float.

EDIT: As commenters pointed out, the explicit cast is only needed in C++, not in C.

Johann Gerell
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