字符*和strcpy的的char []之间的差值（） [英] difference between char* and char[] with strcpy()

问题描述

我一直有麻烦的我，虽然我理解的问题，在过去的几个小时。这里是我的麻烦：

I've been having trouble the past couple hours on a problem I though I understood. Here's my trouble:

void cut_str(char* entry, int offset) {
    strcpy(entry, entry + offset);
}

char  works[128] = "example1\0";
char* doesnt = "example2\0";

printf("output:\n");

cut_str(works, 2);
printf("%s\n", works);

cut_str(doesnt, 2);
printf("%s\n", doesnt);

// output:
// ample1
// Segmentation: fault

我觉得有一些事情的char * /的char []，我不是到这儿。

I feel like there's something important about char*/char[] that I'm not getting here.

推荐答案

的区别在于犯规点属于一个字符串常量的内存，因此是不可写的。

The difference is in that doesnt points to memory that belongs to a string constant, and is therefore not writable.

当你做到这一点。

char  works[128] = "example1\0";

，编译器的拷贝的一个不可写的字符串的内容到一个可写的阵列。 \\ 0 不是必需的，顺便说一句。

the compiler copies the content of a non-writable string into a writable array. \0 is not required, by the way.

当你做到这一点，但是，

When you do this, however,

char* doesnt = "example2\0";

的编译离开指针指向一个不可写的存储器区域。同样， \\ 0 将被编译器插入。

如果您使用的是 GCC ，你可以把它提醒你有关初始化写的char * 与字符串。该选项 -Wwrite串。你会得到像这样的警告：

If you are using gcc, you can have it warn you about initializing writable char * with string literals. The option is -Wwrite-strings. You will get a warning that looks like this:

 warning: initialization discards qualifiers from pointer target type

来声明犯规指针的正确方法如下：

const char* doesnt = "example2\0";

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