从GCC编译C程序段错误：连接两个字符串 [英] c program segment error from GCC compiler: concatenate two strings

问题描述

以下code加薪段误差从 GCC 编译器：

The following code raise segment error from GCC compiler:

char *str1 = "India";
char *str2 = "BIX";
char *str3;
str3 = strcat(str1, str2);
printf("%s %s\n", str3, str1);

问题是 = STR3的strcat（STR1，STR2）。

所以我的问题是：对于GCC编译，它禁止两个指针指向同一件事

So my question is: for GCC Compiler, it forbids two pointers pointing to the same thing?

推荐答案

您应该写什么，code-明智的，回答你的问题：

What you should have written, code-wise, answers your question:

char *str1 = "India";
char *str2 = "BIX";
//should have been
const char *str1 = "India";
const char *str2 = "BIX";

变量 STR1 和 STR2 在内存中的字符串常量印度和 BIX 的位置。该内存的只读的。您尝试使用 strcat的来连接这两个字符串。好吧，看看它做什么：

The variables str1 and str2 are being assigned the address in memory where the string CONSTANTS India and BIX are located. This memory is read-only. You are trying to concatenate these two strings using strcat. Ok, look at what it does:

strcat(str1, str2);
       /\   take this
       || copy  ||
       ===========
        add to end

这需要 STR1 指向的内存足够大，以容纳两个本身就是一个块，你到串联它字符。在字符串的情况下，这是从来没有的情况。结果
还记得我是怎么写你的code与常量存储类说明？现在看看 strcat的的原型：

This requires str1 to point to a block of memory big enough to accomodate both itself and the chars that you're concatenating onto it. In case of string literals, this is never the case.
Also remember how I wrote your code with the const storage class specifier? Now look at strcat's prototype:

char * strcat(char * destination, const char *source);

函数需要的第一个参数是一个普通的的char * ，而不是一个常量。这告诉你的功能将尝试指针指向改变内存。正如其他人所指出的那样，我上面提到的：记忆你的 STR1 来IS的只读指针指向的，因此段错误

The function expects the first argument to be a regular char *, not a const. This tells you the function will attempt to change the memory the pointer points to. As others have pointed out, and I mentioned above: the memory your str1 pointer points to is read only, hence the segfault.

什么你可以写为：

char str1[15] = "India";
const char *str2 = "BIX";
strcat(str1, str2);

这应该工作。需要注意的是 strcat的不检查/ prevent溢出。目标内存的大小应该足够大，以适应的strlen（源）+ strlen的（目标）+ 1 。使用 strncat函数而不是效果会更好，仍然！它允许您指定要连接到目标串字符的最大数量：

That ought to work. Note that strcat does not check/prevent overflow. The size of the destination memory should be big enough to accomodate strlen(source) + strlen(destination) + 1. Using strncat instead would be better, still! It allows you to specify a max number of chars to be concatenated onto the destination string:

size_t free_space = 15;
char str1[free_space] = "India";
const char *str2 = "BIX";
free_space -= strlen(str1) + 1;//add 1 here, to ensure we'll have room for the terminating \0 char
char *str3 = strncat(
    str1,
    str2,
    free_space
);
//now we've concatenated something onto our string, keep track of free space left:
free_space -= strlen(str2);//dangerous, though -> size_t is unsigned, perhaps best do:

size_t src_len = strlen(str2);
char *str3 = strncat(
    str1,
    str2,
    free_space
);
if (src_len >= free_space)
    free_space = 0;
else
    free_space -= src_len;

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