我正在创建四个函数：
就像您将在代码中看到的那样，我使用$(document).on('click', '.card-link', function(){
$(this).next('.card-content').toggle();
});
来查看生成的字符
问题是，排序功能不正常，我无法理解问题出在哪里（我得到一个输出，其中字符不像它们那样排序）。
有人可以告诉我我做错了什么吗？关于如何改进所有代码的任何其他提示也将受到欢迎。
printf答案 0 :(得分：1)
第38行有一个实现错误。下面的详细输出显示了它出错的行。 “ASan”工具声称您拥有heap buffer overflow。
当i为19时，您取消引用p_char[20]，这超出了分配的范围。
$ clang -fsanitize=address -g -Wall sorter.c
sorter.c:11:1: warning: return type of 'main' is not 'int' [-Wmain-return-type]
void main() {
^
sorter.c:11:1: note: change return type to 'int'
void main() {
^~~~
int
1 warning generated.
$ ./a.out
H W J T R H K M J N C T C T L W M S E Q
=================================================================
==7176==ERROR: AddressSanitizer: heap-buffer-overflow on address 0x60300000eff4 at pc 0x0000004cd8af bp 0x7fff8db88420 sp 0x7fff8db88418
READ of size 1 at 0x60300000eff4 thread T0
#0 0x4cd8ae in sort /home/brian/src/so/sorter.c:38:33
#1 0x4cd5e4 in main /home/brian/src/so/sorter.c:15:5
#2 0x7fbeb021da3f in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x20a3f)
#3 0x417508 in _start (/home/brian/src/so/a.out+0x417508)
0x60300000eff4 is located 0 bytes to the right of 20-byte region [0x60300000efe0,0x60300000eff4)
allocated by thread T0 here:
#0 0x4a626b in __interceptor_malloc /home/development/llvm/3.7.0/final/llvm.src/projects/compiler-rt/lib/asan/asan_malloc_linux.cc:40:3
#1 0x4cd65c in allocate /home/brian/src/so/sorter.c:22:22
#2 0x4cd576 in main /home/brian/src/so/sorter.c:13:5
#3 0x7fbeb021da3f in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x20a3f)
SUMMARY: AddressSanitizer: heap-buffer-overflow /home/brian/src/so/sorter.c:38:33 in sort
Shadow bytes around the buggy address:
0x0c067fff9da0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c067fff9db0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c067fff9dc0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c067fff9dd0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c067fff9de0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
=>0x0c067fff9df0: fa fa fa fa fa fa fa fa fa fa fa fa 00 00[04]fa
0x0c067fff9e00: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c067fff9e10: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c067fff9e20: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c067fff9e30: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0c067fff9e40: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
Shadow byte legend (one shadow byte represents 8 application bytes):
Addressable: 00
Partially addressable: 01 02 03 04 05 06 07
Heap left redzone: fa
Heap right redzone: fb
Freed heap region: fd
Stack left redzone: f1
Stack mid redzone: f2
Stack right redzone: f3
Stack partial redzone: f4
Stack after return: f5
Stack use after scope: f8
Global redzone: f9
Global init order: f6
Poisoned by user: f7
Container overflow: fc
Array cookie: ac
Intra object redzone: bb
ASan internal: fe
Left alloca redzone: ca
Right alloca redzone: cb
==7176==ABORTING
答案 1 :(得分：1)
在这里，我使用以下方法更新了sort函数;
/* new sort */
void sort2(char* p_char) {
counter = 0;
for (int i = 0; i < size - 1; i++) {
for (int j = i + 1; j < size; j++) {
if (*(p_char + i) > *(p_char + j)) {
char tmp = *(p_char + i);
*(p_char + i) = *(p_char + j);
*(p_char + j) = tmp;
counter++;
}
}
}
printf("Sort2 process # : %d\n", counter);
}
使用此更新排序，改进了查找解决方案的流程编号，输出如下所示，为50,000个随机字符;
D:\SOF>gcc main.c -Wall -Wextra -pedantic -std=c11 -o main
D:\SOF>main
Sort process # : 598810451
Sort2 process # : 574789
请查看下面的测试代码;
#include <stdio.h>
#include <stdlib.h>
#define MAX_SIZE 50000
int counter = 0;
int size = MAX_SIZE;
char* pArrChar1;
void allocate(char**);
void fill(char*);
void sort(char*);
void sort2(char* p_char); /* updated sort */
void printArr(char*);
int main() {
allocate(&pArrChar1);
fill(pArrChar1);
sort(pArrChar1);
//printArr(pArrChar1);
allocate(&pArrChar1);
fill(pArrChar1);
sort2(pArrChar1);
system("pause");
return 0;
}
void allocate(char** p_char) {
*p_char = (char*)malloc(size*sizeof(char));
}
void fill(char* p_char) {
int max = 90 ;
int min = 65;
for (int i = 0; i < size; i++) {
p_char[i]= (char)(rand() % (min + 1 - max) + min);
//printf("%c ", p_char[i]);
}
//printf("\n\n");
}
void sort(char* p_char) {
counter = 0;
for (int i = 0; i < size; i++) {
for (int j = 0; j < size - 1; j++) {
if (*(p_char + j) > *(p_char + j + 1)) {
char tmp = *(p_char + j);
*(p_char + j) = *(p_char + j + 1);
*(p_char + j + 1) = tmp;
counter++;
}
}
}
printf("Sort process # : %d\n", counter);
}
/* new sort */
void sort2(char* p_char) {
counter = 0;
for (int i = 0; i < size - 1; i++) {
for (int j = i + 1; j < size; j++) {
if (*(p_char + i) > *(p_char + j)) {
char tmp = *(p_char + i);
*(p_char + i) = *(p_char + j);
*(p_char + j) = tmp;
counter++;
}
}
}
printf("Sort2 process # : %d\n", counter);
}
void printArr(char* p_char) {
for (int i = 0; i < size; i++)
printf("%c ", p_char[i]);
printf("\n\n");
}
希望它有所帮助！
答案 2 :(得分：0)
在您的代码中，您有：
for (int i = 0; i < size; i++) {
for (int j = 0; j < size - 1; j++) {
if (*(p_char + i) > *(p_char + i + 1)) {
当i等于size - 1时，您正在访问越界。
更改循环限制：
for (int i = 0; i < size - 1; i++) {
for (int j = 0; j < size; j++) {
if (*(p_char + i) > *(p_char + i + 1)) {
这至少避免了越界数组访问。令人惊讶的是，您没有使用i和j作为下标;你似乎重复了与写作相同的比较。
答案 3 :(得分：0)
sort函数有错误，这是正确的
void sort(char* p_char) {
for (int i = 0; i < size; i++) {
for (int j = 0; j < size - 1; j++) {
if (*(p_char+j) > *(p_char + j + 1)) {
char tmp = *(p_char + j);
*(p_char + j) = *(p_char + j + 1);
*(p_char + j + 1) = tmp;
}
}
}
}