【问题标题】:Getting a seg fault but has no idea how to fix it遇到段错误但不知道如何解决
【发布时间】:2019-09-10 07:32:41
【问题描述】:
以下代码有段错误,但我真的不知道如何调试它,可能是因为我缺乏C语法知识,并且我已经阅读了TCPL但仍然无能为力。
#include <stdio.h>
#include <ctype.h>
int main() {
char *str[4];
char c[2];
for (int i = 0; i < 4; i++)
scanf("%s", str[i]);
int find = 0;
while (find <= 2 && *str[0] != '\0' && *str[1] != '\0') {
if (isalpha(*str[0]) && *str[0] == *str[1]
&& *str[0] - 'A' >= 0 && *str[0] - 'A' <= 25) {
find++;
if (find == 1)
c[0] = *str[0];
else if (find == 2)
c[1] = *str[0];
}
str[0]++;
str[1]++;
}
/* ... */
}
【问题讨论】:
标签:
c
arrays
pointers
debugging
segmentation-fault
【解决方案1】:
这里
char *str[4]; /* what str[0] contains ? some junk data, need to assign valid address */
for (int i = 0; i < 4; i++)
scanf("%s", str[i]); /* No memory for str[i] here */
str 是 字符指针数组 并且它们未初始化,即未指向任何有效地址。解决这个问题的一种方法是为每个字符指针分配内存,然后你可以将一些数据放入str[i]。例如
char *str[4];
for (int i = 0; i < 4; i++) {
str[i] = malloc(MAX); /* define MAX value as per requirement */
scanf("%s", str[i]); /* Now str[i] has valid memory */
}
一旦使用动态内存完成工作,不要忘记通过为每个字符指针调用free(str[i])来释放动态内存以避免内存泄漏。
【解决方案2】:
您忘记了为字符串分配的内存。
您的代码具有动态分配的内存。
#include <stdio.h>
#include <ctype.h>
#include <stdlib.h> //needed for malloc and free
int main() {
char *str[4];
//allocate memory
for (int i = 0; i < 4; ++i) {
//allocate 128B per string
str[i] =(char*) malloc(128 * sizeof(char));
//here you should check if malloc was succesfull
//if malloc failed you schould free previously allocated memory
}
char c[2];
for (int i = 0; i < 4; i++)
scanf("%s", str[i]);
int find = 0;
while (find <= 2 && *str[0] != '\0' && *str[1] != '\0') {
if (isalpha(*str[0]) && *str[0] == *str[1]
&& *str[0] - 'A' >= 0 && *str[0] - 'A' <= 25) {
find++;
if (find == 1)
c[0] = *str[0];
else if (find == 2)
c[1] = *str[0];
}
str[0]++;
str[1]++;
}
//delete memory
for (int i =0; i < 4; ++i) {
free(str[i]);
}
/* ... */
}