是否需要将free()参数强制转换为void *? [英] Is casting free() argument to void * neccessary?

问题描述

是否有必要将此代码段中传递给free()的值转换为void指针?

Is it neccessary to cast the value passed to free() to a void pointer in this code snippet?

free((void *) np->defn);

np是链接列表中的struct，而defn是char *.

np is a struct in a linked list and defn is a char *.

推荐答案

C11:7.23.2.3免费功能

简介

#include <stdlib.h>
void free(void *ptr);  

这意味着它可以将指针带到任何类型(void *是通用指针类型).通常，无需将参数强制转换为void *，但在类似

It means that it can take pointer to any type (void * is a generic pointer type). In general no need to cast the argument to void *, but in cases like

int const *a = malloc(sizeof(int));
free(a);

编译器将生成警告

test.c: In function 'main':
test.c:33:7: warning: passing argument 1 of 'free' discards 'const' qualifier from pointer target type [-Wdiscarded-qualifiers]
  free(a);
       ^
In file included from test.c:2:0:
/usr/include/stdlib.h:151:7: note: expected 'void *' but argument is of type 'const int *'
 void  free(void *);
       ^

要取消显示此警告，需要强制转换

To suppress this warning a cast is needed

 free((void *)a);

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