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程序报段错误_markdown编辑

程序报段错误_markdown编辑

作者: 荣子3507 | 来源:发表于2017-12-01 00:53 被阅读0次

    在代码第51行 *str = *next 处报错:Program received signal SIGSEGV, Segmentation fault

    描述如下:

    编写一个函数,删除一个字符串的一部分。函数原型如下:

    int del_substr (char *str, char const *substr)

    举个例子,假定str指向ABCDEFG. 如果substr指向FGH, CDF或XABC, 函数应该返回0,str未做任何修改。但如果substr指向CDE, 函数就把str修改为指向ABFG.

    代码实现如下:

    #include <stdlib.h>
    #include <stdio.h>
    
    /*
    ** If the string "substr" appears in "str", delete it.
    */
    //#define NULL 0 /* null pointer */
    //#define NUL ’\0’ /* null byte */
    #define TRUE 1
    #define FALSE 0
    /*
    ** See if the substring beginning at ’str’ matches the string ’want’.  If
    ** so, return a pointer to the first character in ’str’ after the match.
    */
    char *match (char *str, char const *want)
    {
        while (*want != '\0') {
            if (*str++ != *want++) {
                return NULL;
            }
        }
        return str;
    }
    
    int del_substr (char *str, char const *substr)
    {
        char *next = NULL;
        char *res = str;
        /*
        ** Look through the string for the first occurrence of the substring.
        */
        while( *str != '\0' ){
           next = match( str, substr );
            if( next != NULL )
                break;
            str++;
        }
        printf("##next: %s\n", next);
        printf("##str: %s\n", str);
        printf("##res: %s\n", res);
        /*
        ** If we reached the end of the string, then the substring was not
        ** found.
        */
        if( *str == '\0' )
            return FALSE;
        /*
        ** Delete the substring by copying the bytes after it over the bytes of
        ** the substring itself.
        */
        while(*next != '\0'){
            *str = *next;
            printf("##str: %s\n", str);
            printf("##next: %s\n", next);
            str++;
            next++;
        }
        *str = '\0';
        printf("after del_substr result: %s\n", res);
        return TRUE;
    }
    
    int main()
    {
        char str[] = "ABCDEFG";
        char const sub_str1[] = "FGH";
        char const sub_str2[] = "DE";
        int res = -1;
       // *str = 'a';
       // *(str+1) = 'b';
        printf("str: %s\n", str);
        res = del_substr(str, sub_str2);
       // printf("the resust is %s\n", str);
        printf("the res is %d\n", res);
        return res;
    }
    

    gdb调试报错如下

    38        printf("##next: %s\n", next);
    (gdb)
    ##next: FG
    39        printf("##str: %s\n", str);
    (gdb)
    ##str: DEFG
    44        if( *str == '\0' )
    (gdb)
    50        while(*next != '\0'){
    (gdb)
    51            (*str) = (*next);
    (gdb)
    Program received signal SIGSEGV, Segmentation fault.
    0x0804849f in del_substr (str=0x8048667 "DEFG", substr=0x8048670 "DE")
    at del_substr.c:51
    51            (*str) = (*next);
    (gdb)
    

    --------------------------------分割线--------------------------------
    知道问题出在哪里了
    main函数最开始的变量初始化
    char *str = "ABCDEFG";
    str指向的是只读常量区,所以执行到 *str = *next;会报错,该语句试图改变只读常量区里的值时,操作系统向程序发送了一个信号,告诉程序操作系统检测到了非法的内存访问,为了防止内存空间被破坏,操作系统提前终止了该程序的运行;
    用数组声明即可
    char str[] = "ABCDEFG";

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