strtok的分段错误 [英] strtok segmentation fault

问题描述

我试图理解为什么code的片断是给分段错误：

I am trying to understand why the following snippet of code is giving a segmentation fault:

void tokenize(char* line)
{
   char* cmd = strtok(line," ");

   while (cmd != NULL)
   {
        printf ("%s\n",cmd);
        cmd = strtok(NULL, " ");
   } 
}

int main(void)
{
   tokenize("this is a test");
}

我知道的strtok（）实际上并不记号化的字符串，但在这种情况下，行直接指向字符串这是一个测试这是内部字符的数组。是否有任何标记化的行没有将其复制到一个数组中？

I know that strtok() does not actually tokenize on string literals, but in this case, line points directly to the string "this is a test" which is internally an array of char. Is there any of tokenizing line without copying it into an array?

推荐答案

的问题是，你正在试图修改字符串。这样做会导致你的程序的行为是不确定的。

The problem is that you're attempting to modify a string literal. Doing so causes your program's behavior to be undefined.

说，你不允许修改字符串是过于简单化。说字符串是常量是不正确;他们不是。

Saying that you're not allowed to modify a string literal is an oversimplification. Saying that string literals are const is incorrect; they're not.

警告：题外话如下

的字符串这是一个测试是类型的前pression的的char [15] （14的长度，再加上1终止'\\ 0'）。在大多数情况下，包括这一个，这样的前pression被隐式转换为指针数组类型的的char * 的第一个元素。

The string literal "this is a test" is of an expression of type char[15] (14 for the length, plus 1 for the terminating '\0'). In most contexts, including this one, such an expression is implicitly converted to a pointer to the first element of the array, of type char*.

试图修改一个字符串文字中提到的阵列的行为是不确定的 - 不是因为它的常量（它不是），而是因为C标准具体说，这是不确定的。

The behavior of attempting to modify the array referred to by a string literal is undefined -- not because it's const (it isn't), but because the C standard specifically says that it's undefined.

有些编译器可能允许您脱身与此有关。您code实际上可能会修改相应的文字（这样会造成很大的混乱以后）的静态数组。

Some compilers might permit you to get away with this. Your code might actually modify the static array corresponding to the literal (which could cause great confusion later on).

大多数现代编译器，虽然，将存储在只读存储器阵列 - 不是物理ROM中，但在是由虚拟存储器系统保护修改的内存区域。试图修改这种存储器的结果通常是分段错误和程序崩溃。

Most modern compilers, though, will store the array in read-only memory -- not physical ROM, but in a region of memory that's protected from modification by the virtual memory system. The result of attempting to modify such memory is typically a segmentation fault and a program crash.

那么，为什么的不的字符串常量？既然你真的不应该试图修改它们，这肯定会是有意义 - 和C ++确实让字符串常量。原因是历史性的。它是由1989年的ANSI C标准出台之前的常量关键字不存在（虽然它由一些编译器在这之前可能实现）。因此，一个pre-ANSI程序可能是这样的：

So why aren't string literals const? Since you really shouldn't try to modify them, it would certainly make sense -- and C++ does make string literals const. The reason is historical. The const keyword didn't exist before it was introduced by the 1989 ANSI C standard (though it was probably implemented by some compilers before that). So a pre-ANSI program might look like this:

#include <stdio.h>

print_string(s)
char *s;
{
    printf("%s\n", s);
}

main()
{
    print_string("Hello, world");
}

有没有办法强制执行 print_string 不允许修改字符串指向的事实小号。制作字符串常量在ANSI C将会破坏现有的code，它的ANSI C委员会极力避免这样做。没有发生过从那时起一个很好的机会来做出这样的改变的语言。 （C ++，主要是Bjarne的Stroustrup的，的设计师们并不像担心用C的向后兼容性。）

There was no way to enforce the fact that print_string isn't allowed to modify the string pointed to by s. Making string literals const in ANSI C would have broken existing code, which the ANSI C committee tried very hard to avoid doing. There hasn't been a good opportunity since then to make such a change to the language. (The designers of C++, mostly Bjarne Stroustrup, weren't as concerned about backward compatibility with C.)

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