c - 为什么我的 C 代码每次都不起作用？

Question

我编写这段代码是为了合并两个已排序的数组。期望的输出是:

Merged array:0 1 2 3 4 5 6 7 8 9 10 11 12

我正在使用 gcc (Ubuntu 5.4.0-6ubuntu1~16.04.4) 5.4.0 20160609 来编译我的代码。

问题是有时我在执行 a.out 文件时会得到所需的输出，但在其他情况下，光标会一直闪烁并且没有显示任何结果。为什么会这样？我的代码有问题吗？

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

int main(void){

//change arrays as per your need but it should be sorted
int a[] = {1,2,3,7,8};
int b[] = {0,3,5,6,9,10,11,12};

int m =sizeof(a) / sizeof(int);
int n =sizeof(b) / sizeof(int);

int index=0, j=0, k=0;
int size = m + n;
int array[size];


while(index < size) {

    while(a[j] < b[k] && j<m ){
        array[index] = a[j];
        ++index;
        ++j;
    }
    while(a[j] > b[k] && k<n){
        array[index] = b[k];
        ++index;
        ++k;            
    }
    while(a[j] == b[k]){
        array[index] = a[j];
        j++; index++;            
    }        
}

printf("Merged array: ");
for(int i=0; i<size; i++)
    printf("%d ", array[i]);

printf("\n");

}

Answer 1

您有未定义的行为(越界访问数组)。使用 gcc -fsanitize=undefined 创建一个可以检测各种不良行为的可执行文件。

% gcc -g fffff.c -Wall -Wextra -fsanitize=undefined
% ./a.out
fffff.c:20:12: runtime error: index 5 out of bounds for type 'int [5]'
fffff.c:20:12: runtime error: load of address 0x7ffd0c0c9804 with insufficient space for an object of type 'int'
0x7ffd0c0c9804: note: pointer points here
  08 00 00 00 04 00 00 00  04 00 00 00 04 00 00 00  00 00 00 00 03 00 00 00  05 00 00 00 06 00 00 00
              ^ 
fffff.c:25:12: runtime error: index 5 out of bounds for type 'int [5]'
fffff.c:25:12: runtime error: load of address 0x7ffd0c0c9804 with insufficient space for an object of type 'int'
0x7ffd0c0c9804: note: pointer points here
  08 00 00 00 04 00 00 00  04 00 00 00 04 00 00 00  00 00 00 00 03 00 00 00  05 00 00 00 06 00 00 00
              ^ 
fffff.c:30:12: runtime T: index 5 out of bounds for type 'int [5]'
fffff.c:30:12: runtime error: load of address 0x7ffd0c0c9804 with insufficient space for an object of type 'int'
0x7ffd0c0c9804: note: pointer points here
  08 00 00 00 04 00 00 00

第20、25、30行是

20      while(a[j] < b[k] && j<m ){

25      while(a[j] > b[k] && k<n){

30      while(a[j] == b[k]){