Question

我有以下方案试图证明(基于空间的)扼杀,并把它写成一个果园。

#include <stdio.h>
#include <stdlib.h>

#define len 180

void tokenize(char *str, char **tokens)
{
    int l = 0, index = 0;
    int i = 0;
    int str_i;
    int tok_i;

    while(*str) {
        if (*str ==    ) {
            tokens[i] = malloc(sizeof(char) * l+1);
            if (tokens[i] == NULL) return;

            tok_i = 0;           
            for (str_i=index-len ; str_i<index ; str_i++) {
                tokens[i][tok_i] = str[str_i];
                tok_i++;
            }

            tokens[i][tok_i] =   ;
            tokens[i++] = NULL;
            l = 0;
            index++;
        }
        str++;
        l++;
        index++;
    }       

    return;         
}

int main()
{
    char str[len] = "this is a test string";
    char **tokens = malloc(100 * sizeof(char *));

    if (str == NULL || tokens == NULL)
        return 1;

    printf("input string: %s
", str);
    tokenize(str, tokens);

    return 0;
}

上述方案汇编了罚款,但在执行时,我在malloc.c上作了以下陈述:

$ gcc -ggdb -Wall prog.c 
$ ./a.out 
input string: this is a test string
a.out: malloc.c:2453: 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)
$

st痕表明:

(gdb) bt
#0  0x0000003b28036285 in raise () from /lib64/libc.so.6
#1  0x0000003b28037b9b in abort () from /lib64/libc.so.6
#2  0x0000003b2807d37d in __malloc_assert () from /lib64/libc.so.6
#3  0x0000003b28080c37 in _int_malloc () from /lib64/libc.so.6
#4  0x0000003b28082595 in malloc () from /lib64/libc.so.6
#5  0x000000000040055f in tokenize (str=0x7fffffffe017 " a test string", tokens=0x601010) at prog.c:15
#6  0x00000000004006de in main () at prog.c:46
(gdb)

我如何去掉这一点? 任何想法都会受到高度赞赏。

Answer 1

页: 1

for (str_i=index-len ; str_i<index ; str_i++)

str_i是负数。

Answer 2

for (str_i=index-len ; str_i<index ; str_i++) {
        tokens[i][tok_i] = str[str_i];
        tok_i++;
}

tokens[i][tok_i] =   ;

至少其中之一的<代码>表示[i][tok_i]是指在分配的记忆外的,以便你最终放弃malloc。之后,hilarity subsequentlyes 。

<代码>str_i的初始化也令人感兴趣,因为第一个版本的编号为-180。

Answer 3

是的,根据@cnicutar和@spicavigo的建议,我修改了该法典如下,并做了罚款。

#include <stdio.h>
#include <stdlib.h>

#define str_len 180

void tokenize(char *str, char **tokens)
{
    int length = 0, index = 0;
    int i = 0;
    int str_i;
    int tok_i;

    while(str[length]) {
        if (str[length] ==    ) {
            /* this charecter is a space, so skip it! */
            length++;
            index++;

            tokens[i] = malloc(sizeof(char) * (index+1));

            tok_i = 0;           
            for (str_i=length-index ; str_i<length; str_i++) {
                tokens[i][tok_i] = str[str_i];
                tok_i++;
            }

            tokens[i][tok_i] =   ;
            i++;
            index = 0;
        }
        length++;
        index++;
    }       

    /* copy the last word in the string */
    tokens[i] = malloc(sizeof(char) * index);
    tok_i = 0;           
    for (str_i=length-index ; str_i<length; str_i++) {
        tokens[i][tok_i] = str[str_i];
        tok_i++;
    }
    tokens[i][tok_i] =   ;
    tokens[i++] = NULL;

    return;         
}

int main()
{
    char *str = malloc(str_len * sizeof(char));
    char **tokens = malloc(100 * sizeof(char *));
    int i = 0;

    if (str == NULL || tokens == NULL)
        return 1;

    gets(str);
    printf("input string: %s
", str);
    tokenize(str, tokens);

    while(tokens[i] != NULL) {
        printf("%d - %s 
", i, tokens[i]);
        i++;
    }

    while(tokens[i])
        free(tokens[i]);
    free(tokens);
    free(str);

    return 0;
}

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