未初始化的值是由堆栈分配创建的 - valgrind

Sum*_*mit 0 c valgrind

我用valgrind用选项调试我的代码track-origins=yes并遇到了这个错误.

$ valgrind --track-origins=yes ./frgtnlng < in > out
==7098== 
==7098== Conditional jump or move depends on uninitialised value(s)
==7098==    at 0x4C2F1BC: strcmp (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==7098==    by 0x400857: main (frgtnlng.c:24)
==7098==  Uninitialised value was created by a stack allocation
==7098==    at 0x40064C: main (frgtnlng.c:9)
==7098== 
==7098== Conditional jump or move depends on uninitialised value(s)
==7098==    at 0x40085A: main (frgtnlng.c:24)
==7098==  Uninitialised value was created by a stack allocation
==7098==    at 0x40064C: main (frgtnlng.c:9)
Run Code Online (Sandbox Code Playgroud)

第9行是:

scanf("%d", &t);
Run Code Online (Sandbox Code Playgroud)

我不明白这是如何导致问题的.

frgtnlng.c:

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

int main(void)
{
    int t, n, k, l, i, j, z, out[100];
    char f[5][100], m[5][50][50];

    scanf("%d", &t);
    while (t--) {
        for (i = 0; i < 100; i++)
            out[i] = 0;
        scanf("%d%d", &n, &k);
        for (i = 0; i < n; i++)
            scanf("%s", f[i]);
        for (i = 0; i < k; i++) {
            scanf("%d", &l);
            for (j = 0; j < l; j++)
                scanf("%s", m[i][j]);
        }
        for (i = 0; i < k; i++)
            for (j = 0; j < l; j++)
                for (z = 0; z < n; z++) {
                    if (strcmp(m[i][j], f[z]) == 0)
                        out[z] = 1;
                }
        for (i = 0; i < n; i++) {
            if (out[i])
                printf("YES ");
            else
                printf("NO ");
        }
        printf("\n");
    }
    return 0;
}
Run Code Online (Sandbox Code Playgroud)

在:

2
3 2
piygu ezyfo rzotm
1 piygu
6 tefwz tefwz piygu ezyfo tefwz piygu
4 1
kssdy tjzhy ljzym kegqz
4 kegqz kegqz kegqz vxvyj
Run Code Online (Sandbox Code Playgroud)

das*_*ght 7

valgrind的行号是关闭的:它应该为分配行号报告7而不是9.然而,错误行24是正确的 - 问题出在这里:

if (strcmp(m[i][j], f[z]) == 0)
Run Code Online (Sandbox Code Playgroud)

问题是j从0到l-1包含的循环,但l它是在读取2D数组的循环的最后一次迭代中设置的任何东西,即4.这就是为什么每次它到达数组中的行少于它从数组的未初始化部分读取的4个条目.

修复是通过创建l一个数组来单独存储各行的长度l[5],并l[i]在两个循环中使用:

for (i = 0; i < k; i++) {
    scanf("%d", &l[i]);
    for (j = 0; j < l[i]; j++)
        scanf("%s", m[i][j]);
}
for (i = 0; i < k; i++)
    for (j = 0; j < l[i]; j++)
        for (z = 0; z < n; z++) {
            if (strcmp(m[i][j], f[z]) == 0)
                out[z] = 1;
        }
Run Code Online (Sandbox Code Playgroud)