解决“invalid conversion from void* to char*”和“deprecated conversion from string constant to 'char*' "
2017-07-01 07:49
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先看下面这个C语言程序,它的作用是使用动态存储分配来创建字符串的副本,如果内存分配失败,
使用
但当使用
error: invalid conversion from ‘void*’ to ‘char*’ [-fpermissive]
warning: deprecated conversion from string constant to ‘char*’ [-Wwrite-strings]
出现error的原因是C++设计得比C更加安全,它不能自动地将
duplicate函数就返回空指针。
#include <stdio.h> #include <stdlib.h> #include <string.h> char* duplicate(const char* s) { char* temp = malloc(strlen(s) + 1); if (temp == NULL) return NULL; strcpy(temp, s); return temp; } int main(void) { char* str = "testing."; char* str2 = duplicate(str); printf("Original String: %s\n", str); printf("Duplicated String: %s\n", str2); }
使用
gcc编译直接通过并打印出下面的结果
Original String: testing. Duplicated String: testing.
但当使用
g++编译时,会出现一个错误和警告,如下
error: invalid conversion from ‘void*’ to ‘char*’ [-fpermissive]
warning: deprecated conversion from string constant to ‘char*’ [-Wwrite-strings]
出现error的原因是C++设计得比C更加安全,它不能自动地将
void *转换为其它指针类型。而出现warning的原因则是程序试图将字符串字面值(在C++中字符串字面值为const char []类型,而在C语言中则为char []类型)转换为
char *类型,因此如果想要使用
g++成功编译这个程序并得到预期的结果,可以将源程序修改为
#include <stdio.h> #include <stdlib.h> #include <string.h> char* duplicate(const char* s) { char* temp = (char*) malloc(strlen(s) + 1); if (temp == NULL) return NULL; strcpy(te 96c5 mp, s); return temp; } int main(void) { char* str = (char*) "testing."; char* str2 = duplicate(str); printf("Original String: %s\n", str); printf("Duplicated String: %s\n", str2); }
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