奇怪的错误：中止陷阱，同时处理字符数组用C [英] Weird Error: Abort trap while handling character array in C

问题描述

欲存储字符串中的每个字符的二进制值，并将其存储在数组中。但是，当我开始瞎搞像memset的功能，我有过调试无法控制。

I want to store the binary value of each character in the string and store it in an array. But when i start messing around with functions like memset, i have no control over the debugging.

#include <stdio.h>
#include <string.h>
int main()
{
char str[8];
char *ptr = "Hello";
int i;

for(; *ptr != 0; ++ptr)
{
    printf("%c => ", *ptr);

    /* perform bitwise AND for every bit of the character */
    for(i = 7; i >= 0; --i) 
        if(*ptr & 1 << i) 
          str[7-i]='1';
        else
          str[7-i]='0';
        //(*ptr & 1 << i) ? putchar('1') : putchar('0');
    str[8]='\0';    
    printf("%s\n",str);
    memset(str,'/0',8);        
}
return 0;
}

输出：
H => 01001000
E => 01100101
L => 01101100
L => 01101100
O => 01101111
中止陷阱

Output: H => 01001000 e => 01100101 l => 01101100 l => 01101100 o => 01101111 Abort trap

这将是很好，如果有人可以抛出一些轻。即使我得到的输出，陷阱是发生在手。

It would be nice if someone can throw some light. Even though i am getting the output, the trap is happening in the hand.

礼貌：这是一个老乡的修改程序栈的同胞用户@Athabaska

Courtesy: This is a modified program of a fellow stack fellow user @Athabaska.

谢谢！

推荐答案

海峡[8] ='\\ 0'导致缓冲区溢出，因为 STR 只能有8个元素，而你试图访问第九之一。

str[8]='\0' causes a buffer overflow because str only has 8 elements, and you're trying to access the ninth one.

编辑：我抱怨'/ 0'，但作为@R ..告诉我，它的是的有效的 - 但是， memset的（）最终将其参数转换为 unsigned char型，所以我不认为有任何一点在使用它，所以我想笔者希望将其更改为'\\ 0'。

I complained about the '/0', but as @R.. informed me, it is valid - however, memset() will ultimately convert its parameter to unsigned char, so I don't think there is any point in using it, so I assume the author will want to change it to '\0'.

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