我试图从stdin中标记字符串。我只对字符感兴趣,并为每个字符串构建一个数组(忽略非字符)。出于某种原因,当我从stdin中读取24个或更多字符时,收到错误:
free():下一个大小(fast)无效:
这是相关的代码…它在较小的字符串(23个字符或更少)上运行良好
char **tokenize(int *nbr_words) {
char **list = calloc(INITIAL_SIZE, sizeof(char *));
char * temp = NULL;
temp = malloc(sizeof(200));
while(fgets(temp,200,stdin)){
char * newWord = NULL;
newWord = malloc(sizeof(100));
int i = 0;
while(temp[i] != '\n'){
if(isalpha(temp[i]) && temp[i+1] != '\n'){
strncat(newWord,&temp[i],1);
i++;
}
else if(isalpha(temp[i]) && temp [i+1] == '\n'){
strncat(newWord,&temp[i],1);
list[*nbr_words] = newWord;
*nbr_words += 1;
printf("%s\n",list[*nbr_words -1]);
i++;
if(*nbr_words % 10 == 9){
list = realloc(list, *nbr_words + 10);
}
free(newWord);
newWord = malloc(sizeof(100));
*newWord = NULL;
}else{
if(*newWord == NULL){
i++;
}
else if(*nbr_words % 10 != 9){
list[*nbr_words] = newWord;
*nbr_words += 1;
printf("%s\n",list[*nbr_words-1]);
i++;
free(newWord);
newWord = malloc(sizeof(100));
*newWord = NULL;
}else{
list = realloc(list, *nbr_words + 10);
list[*nbr_words] = newWord;
*nbr_words += 1;
printf("%s\n",list[*nbr_words-1]);
i++;
free(newWord);
newWord = malloc(sizeof(100));
*newWord = NULL;
}
}
}
free(temp);
temp = malloc(sizeof(200));
*temp = NULL;
}
return list;
}
最佳答案
您继续为大小temp
或newWord
分配sizeof(200)
和sizeof(100)
,而这两者都等于sizeof(int)
,这比您预期的要小得多。
修改
temp = malloc(sizeof(200));
newWord = malloc(sizeof(100));
进入之内
temp = malloc(200);
newWord = malloc(100);
关于c - free():无效的下一个尺寸(快速)错误,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/19506592/