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c - 这个 c 程序在 windows 中工作正常但在 Linux 中不工作

转载 作者:太空宇宙 更新时间:2023-11-04 11:59:16 24 4
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此 c 程序在 Windows 中运行良好,但在 Linux 中显示段错误。

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

void comb(long int *arr,long int n,long int r,long int stick)
{
long int check=1,sum =0;
int poscheck = 0,status = 0;
long int *temp = malloc(r * sizeof(long int));
long int *pos = malloc(r * sizeof(long int));
long int *rept = malloc(r * sizeof(long int));
memset(pos, 0, r*sizeof(long int));
memset(rept, 0, r*sizeof(long int));

while (check <= pow(n,r))
{
for (long int i = 0; i < r; i++) //for making the number of array
{
for(long int j = 0; j < r; j++) //For checking that no number is repeating.
{
if(i == j) continue; //for skip checking of the same element
else if(pos[i] == pos[j])
{
poscheck = 1;
break;
}
}
if(poscheck == 1) break;
temp[i] = arr[pos[i]];
sum += temp[i];
}
if((sum == stick) && poscheck == 0)
{
for(long int i = 0 ; i< r ; i++)
{
printf("%ld ",temp[i]);
}
status = 1;
printf("\n");
break;
}
sum = 0,poscheck = 0;
for (long int i = 0; i < r; i++)
{
if (pos[i] == n - 1)
{
rept[i]++; //To check how much time the number is repeated in a column
}
if ((pos[i] == n - 1) && (rept[i] == pow(n, r-i-1))) //If it is repeated a specific number of time then change the value of it's previous position
{
if (pos[i - 1] == n - 1) //if the previous number is the last number then it will start the series again
{
pos[i - 1] = 0;
}
else
pos[i - 1]++; //If the previous number is not the last number of series then go to the next number
rept[i] = 0;
}
}
if (pos[r - 1] < n - 1) //for go to the next number of series in the last line
{
pos[r - 1]++;
}
else
{
pos[r - 1] = 0; //if it is the last number of series then start form the first again
}
check++;
}
if(status == 0)
{
printf("-1\n");
}
free(pos); //Does not know why this is showing "double free or corruption (out)" in linux but working in windows.
free(rept);
free(temp);
}
int main()
{
long int n,data[3],j=0;
scanf("%ld",&n);
long int *arr = malloc(n*sizeof(long int));
while(j < n)
{
for(long int i = 0; i< 3; i++)
{
scanf("%ld",&data[i]);
}
for(long int i = 0; i < data[1]; i++)
{
arr[i] = i+1;
}
comb(arr,data[1],data[2],data[0]);
j++;
}
free (arr);
return 0;
}

给定的输入是

12 8 3
10 3 3
9 10 2
9 10 2

这是在 linux 中显示

1 3 8 
-1
munmap_chunk(): invalid pointer
Aborted (core dumped)

这在窗口中显示完美

2 3 7
-1
5 4
1 8

我在 Windows 和 Linux 中都使用了 gcc 和 tcc 编译器,但在 Linux 中都给出了相同的错误。

无法理解为什么问题会出现在 Linux 中。

最佳答案

如果输入是

1 3 8 
-1

主要

n = 1

arr = memory for 1 long int

in for(long int i = 0; i< 3; i++)

data array = 3 8 -1

in: for(long int i = 0; i < data[1]; i++)

arr[i] = some value    

但是 arr 数组将迭代 8 次并尝试分配 arr[i],即使 arr 只有一个元素。

关于c - 这个 c 程序在 windows 中工作正常但在 Linux 中不工作,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53122062/

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