为什么将字符串文字传递给 char* 参数有时只是编译器错误?

2022-01-12 00:00:00 c char constants c++ string-literals

我正在使用 C 和 C++ 程序.我们曾经在没有 make-strings-writable 选项的情况下进行编译.但那会收到一堆警告，所以我把它关掉了.

I'm working in a C, and C++ program. We used to be compiling without the make-strings-writable option. But that was getting a bunch of warnings, so I turned it off.

然后我收到一大堆错误，形式为无法在函数 foo 的参数 3 中将 const char* 转换为 char*".所以，我经历了很多改变来解决这些问题.

Then I got a whole bunch of errors of the form "Cannot convert const char* to char* in argmuent 3 of function foo". So, I went through and made a whole lot of changes to fix those.

但是，今天，程序崩溃了，因为文字"被传递到一个期待 char* 的函数中，并将第 0 个字符设置为 0.它没有做任何坏事，只是试图编辑一个不断的，崩溃的.

However, today, the program CRASHED because the literal "" was getting passed into a function that was expecting a char*, and was setting the 0th character to 0. It wasn't doing anything bad, just trying to edit a constant, and crashing.

我的问题是，为什么这不是编译器错误?

My question is, why wasn't that a compiler error?

如果重要的话，这是在使用 gcc-4.0 编译的 mac 上.

In case it matters, this was on a mac compiled with gcc-4.0.

添加代码:

char * host = FindArgDefault("EMailLinkHost", "");
stripCRLF(linkHost, '
');

地点:

char *FindArgDefault(char *argName, char *defVal) 
{// simplified
    char * val = defVal;
    return(val);
}

和

void stripCRLF(char *str, char delim)
{
    char *p, *q;

    for (p = q = str; *p; ++p) {
        if (*p == 0xd || *p == 0xa) {
            if (p[1] == (*p ^ 7)) ++p;
            if (delim == -1) *p = delim;
            }
        *q++ = *p;
        }
    *q = 0;  // DIES HERE
}

编译并运行，直到它尝试将 *q 设置为 0...

This compiled and ran until it tried to set *q to 0...

编辑 2:

大多数人似乎都没有抓住我的问题的重点.我知道为什么 char foo[] = "bar" 有效.我知道为什么 char * foo = "bar";不起作用.

Most people seem to be missing the point of my question. I know why char foo[] = "bar" works. I know why char * foo = "bar"; doesn't work.

我的问题主要是关于传递参数.我想到的一件事是这可能是 C 与 C++ 的问题吗?"因为我有一些 .c 文件和一些 .cpp 文件，而且 C 很可能允许它，但 C++ 不允许......反之亦然......

My question is mostly with respect to passing parameters. One thing that occures to me is "Is it possible that this is a C vs C++ issue?" because I have some .c files and some .cpp files, and it's quite possible that C allows it, but C++ doesn't... or vice versa...

推荐答案

标准规定了一个特殊的规则，允许字面量到 char* 的转换，它悄悄地丢弃了 const资格.(4.2/2):

The standard specifies a special rule allowing the literal-to-char* conversion which quietly drops const qualification. (4.2/2):

不是宽字符串文字的字符串文字(2.13.4)可以转换为pointer to char"类型的右值；宽字符串文字可以转换为指向 wchar_t 的指针"类型的右值.无论哪种情况，结果都是指向数组第一个元素的指针.仅当存在显式适当的指针目标类型时才考虑这种转换，而不是当一般需要从左值转换为右值时.[注意:此转换已弃用.见附录 D.]

A string literal (2.13.4) that is not a wide string literal can be converted to an rvalue of type "pointer to char"; a wide string literal can be converted to an rvalue of type "pointer to wchar_t". In either case, the result is a pointer to the first element of the array. This conversion is considered only when there is an explicit appropriate pointer target type, and not when there is a general need to convert from an lvalue to an rvalue. [Note: this conversion is deprecated. See Annex D. ]

C++0x 标准进一步弃用了这一点……这条废话规则已从即将发布的标准中完全删除.

The C++0x standard takes that deprecation further… this nonsense rule is removed entirely from the upcoming standard.

const char* 到 char* 错误必须是先将文字转换为 const char* 的结果.

The const char* to char* error must be a result of converting a literal to a const char* first.

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