C++ 将字符串文字分配给 char*
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Assign a string literal to a char*
提问by Pietro M
Possible Duplicate:
How to get rid ofdeprecated conversion from string constant to ‘char*'warnings in GCC?
可能的重复：
如何消除deprecated conversion from string constant to ‘char*'GCC 中的警告？
This assignment:
这个任务：
char *pc1 = "test string";
gives me this warning:
给我这个警告：
warning: deprecated conversion from string constant to 'char*'
警告：不推荐使用从字符串常量到 'char*' 的转换
while this one seems to be fine:
虽然这个似乎没问题：
char *pc2 = (char*)("test string");
Is this one a really better way to proceed?
这是一种更好的方法吗？
Notes: for other reasons I cannot use a const char*.
注意：由于其他原因，我不能使用const char*.
回答by Wyzard
A string literal is a const char[]in C++, and may be stored in read-only memory so your program will crash if you try to modify it. Pointing a non-const pointer at it is a bad idea.
字符串文字是const char[]C++ 中的 a，并且可能存储在只读内存中，因此如果您尝试修改它，您的程序将会崩溃。将非常量指针指向它是一个坏主意。
回答by NPE
In your second example, you must make sure that you don't attempt to modify the the string pointed to by pc2.
在第二个示例中，您必须确保不尝试修改 指向的字符串pc2。
If you do need to modify the string, there are several alternatives:
如果确实需要修改字符串，有几种选择：
Make a dynamically-allocated copy of the literal (don't forget to
free()it when done):char *pc3 = strdup("test string"); /* or malloc() + strcpy() */Use an array instead of a pointer:
char pc4[] = "test string";
制作文字的动态分配副本（完成后不要忘记
free()它）：char *pc3 = strdup("test string"); /* or malloc() + strcpy() */使用数组而不是指针：
char pc4[] = "test string";
回答by Mahesh
That depends on whether you need to modify the string literal or not. If yes,
这取决于您是否需要修改字符串文字。如果是，
char pc1[] = "test string";
