Why must I cast the result from malloc?

The following code gives the error "error: invalid conversion from void* to char* [-fpermissive]"

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

int main(){
    char *t = malloc(20);
}

However, casting the result of malloc solves the problem. But I cannot understand why as this question says that casting the result of malloc is not needed.

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You compiled this C program using a C++ compiler. It is necessary to cast the result of malloc in C++, but not in C.

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Is it possible that you are compiling the code as c++? In c++ the cast would be required.

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In your case,

malloc(20);

will assign memory as required and will return a pointer which can be assigned to any pointer, as explained in this document

As pointed out, you probably are using C++ compiler, because in C its actually considered a bad practice to cast result of malloc.

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