Question

我试图用空格和引号来标记一个单词，但是在输出正确后我得到了一个奇怪的malloc错误。

我希望这个函数接受类似的东西：

hello world "SOme quote"

，输出应为：

hello
world
"some quote"

或输入是：

hello world no quote

输出应为：

hello
world
no
quote

<击> 但是现在，我得到了：

Hello WOrld "Hello World" *** glibc detected *** ./a.out: free(): invalid next size (fast): 0x0000000001760010 *** a.out: malloc.c:2451: sYSMALLOc: Assertion `(old_top == (((mbinptr) (((char *) &((av)->bins[((1) - 1) * 2])) - __builtin_offsetof (struct malloc_chunk, fd)))) && old_size == 0) || ((unsigned long) (old_size) >= (unsigned long)((((__builtin_offsetof (struct malloc_chunk, fd_nextsize))+((2 * (sizeof(size_t))) - 1)) & ~((2 * (sizeof(size_t))) - 1))) && ((old_top)->size & 0x1) && ((unsigned long)old_end & pagemask) == 0)' failed. Aborted (core dumped)

看起来输出是正确的，但是在
后它会变得混乱
代码是：

int process_command(char command[80]){ char curr_char; char *word; int start_pos; int i; int len; len = strlen(command); for(i=0,start_pos=0;i<strlen(command);i++){ curr_char = command[i]; if (curr_char == ' '){ if (command[i-1]==' ') {start_pos++;continue;} word = malloc(i-start_pos*(sizeof(char))); strncpy(word,command+start_pos,i-start_pos); printf("%s\n",word); free(word); start_pos =i+1; } else if (curr_char == '\"'){ word= malloc(len-i*(sizeof(char))); strncpy(word,command+i,len); printf("%s\n",word); free(word); i=len+len; } } return 0; } int main(){ char buffer[80] = "Hello WOrld \"Hello World\""; process_command(buffer); return 0; }

<击>

问题得到解决！谢谢下面是更新后的代码：

int process_command(char command[80]){ char curr_char; char *word; int start_pos; int i; int len; int quote=0; len = strlen(command); for(i=0,start_pos=0;i<strlen(command);i++){ curr_char = command[i]; if (curr_char == ' '){ /*If there was a space found copy the stuff before the space*/ if ( i>0 && command[i-1]==' ') { start_pos++; continue; } word = malloc(i-start_pos+1*(sizeof(char))); strncpy(word,command+start_pos,i-start_pos); word[i-start_pos+1]='\0'; printf("%s\n",word); free(word); start_pos =i+1; } else if (curr_char == '\"'){ /*If a quote was found, copy the rest of the string and exit loop*/ word= malloc(len-i+1*(sizeof(char))); strncpy(word,command+i,len-i); word[len-i+1]='\0'; printf("%s\n",word); free(word); quote=1; break; } }if (quote==0){ /*If there was no quote in the string, get the last element*/ word = malloc(len-start_pos+1*(sizeof (char))); strncpy(word,command+start_pos,len-start_pos); word[len-start_pos+1]='\0'; printf("%s\n",word); free (word); } return 0; } int main(){ char buffer[80] = "Hello \"WOrld test\""; process_command(buffer); return 0; }

但是，我想知道这是否是一种有效的标记方式？这是处理用户键入的指令。因此，如果用户输入

add 1 2 "SOme text"

我想将查询标记为三个部分然后进行处理。所以要做到这一点，我正在对它进行标记，并将它们推入队列，以后我可以逐个弹出项目并处理它。

Answer 1

你必须确保为你要做的strncpy分配足够的内存。这两行是一对一的，因为strncpy也写了一个零字节：

word = malloc(i-start_pos*(sizeof(char)));
strncpy(word,command+start_pos,i-start_pos);

这两行没有意义：你分配len-i字节，然后写len字节（加上零字节）给它：

word = malloc(len-i*(sizeof(char)));
strncpy(word,command+i,len);

Answer 2

您有几个问题，其中一些是：

for(i=0,start_pos=0;i<strlen(command);i++){
    curr_char = command[i];
    if (curr_char == ' '){
        if (command[i-1]==' ') {start_pos++;continue;}      // accesses an invalid array offset when i == 0
        word = malloc(i-start_pos*(sizeof(char)));          // doesn't allocate space for null terminator
        strncpy(word,command+start_pos,i-start_pos);        // doesn't null terminate the string
        printf("%s\n",word);
        free(word);

        start_pos =i+1;

    }
    else if (curr_char == '\"'){
        word= malloc(len-i*(sizeof(char)));                 // doesn't allocate space for null terminator
        strncpy(word,command+i,len);                        // writes past the end of the allocated buffer
        printf("%s\n",word);
        free(word);
        i=len+len;                                          // not sure what the intent of this is?  use `break;`?
    }

}

一般来说，strncpy()应该避免，因为它并不总是按用户的意料行事，所以它经常涉及错误的代码。

此外，您的报价处理非常简单 - 它假定引用的项始终是字符串中的最后一个标记。这可能是你想要的，但它不适用于一组令牌，如：

"one and two" three

C中奇怪的malloc错误，标记单词

2 个答案: