从valgrind读取大小为1的无效

时间:2016-07-21 23:47:44

标签: c memory-leaks valgrind

我修复了内存泄漏后,valgrind向我展示了一堆行错误,如下所示,我不知道如何修复它。是因为我释放的空间比我需要的还要多吗?

第39行:

root = importTree(*(argv+1));

第72行:

 printf("Result found for %d:\n       city: %s\n       state:%s\n", zip, new->city, new->state);

Node *importTree(char *filename) {
    Node *root = NULL;
    FILE *fp = fopen(filename, "r");

    if (!fp) {
        printf("Error opening file.\n");
        return NULL;
    }

    while (!feof(fp)) {
        Node *new = malloc(sizeof(Node));
        if (!new) {
            printf("Failed to allocate memory. Ending read.\n");
            exit(1);
            fclose(fp);
        }
        new->city = malloc(sizeof(char) * MAXCITYNAME);
        if (!(new->city)) {
            printf("Failed to allocate memory. Ending read.\n");
            exit(1);
            fclose(fp);
        }
        new->left = NULL;
        new->right = NULL;
        char *line = malloc(sizeof(char) * MAXLINELENGTH);
        if (!line) {
            printf("Failed to allocate memory. Ending read.\n");
            exit(1);
            fclose(fp);
        }
        if (fgets(line, MAXLINELENGTH, fp) == NULL) {
            if (!feof(fp)) {
                printf("File reading ended prematurely. Check for errors in the file.\n");
                exit(1);
                fclose(fp);
            }
            free(new->city);
            free(line);
            free(new);
            fclose(fp);
            break;
        }
        char *tmp = strtok(line, ",");
        new->zipCode = atoi(tmp);
        tmp = strtok(NULL, ",");
        strcpy(new->city, tmp);
        new->city[strlen(tmp) + 1] = '\0';
        tmp = strtok(NULL, ",");
        strcpy(new->state, tmp);
        new->state[2] = '\0';
        root = addNode(root, new);
        if (!root) {
            printf("Root of tree is still NULL! Ending read.\n");
            exit(1);
        }
        free(line);
        free(new->city); \\this is the line 220
    }
    fclose(fp);
    return root;
}

这是valgrind的输出:

Invalid read of size 1
==7879==    at 0x517FAB4: vfprintf (vfprintf.c:1635)
==7879==    by 0x5188C98: printf (printf.c:34)
==7879==    by 0x400BBD: main (hw3.c:72)
==7879==  Address 0x5657a30 is 0 bytes inside a block of size 30 free'd
==7879==    at 0x4C2AD17: free (in /usr/lib64/valgrind/vgpreload_memcheck-amd64-linux.so)
==7879==    by 0x40103F: importTree (hw3.c:220)
==7879==    by 0x400A31: main (hw3.c:39)
==7879==
==7879== Invalid read of size 1
==7879==    at 0x51ADA99: _IO_file_xsputn@@GLIBC_2.2.5 (fileops.c:1342)
==7879==    by 0x517FA6C: vfprintf (vfprintf.c:1635)
==7879==    by 0x5188C98: printf (printf.c:34)
==7879==    by 0x400BBD: main (hw3.c:72)
==7879==  Address 0x5657a37 is 7 bytes inside a block of size 30 free'd
==7879==    at 0x4C2AD17: free (in /usr/lib64/valgrind/vgpreload_memcheck-amd64-linux.so)
==7879==    by 0x40103F: importTree (hw3.c:220)
==7879==    by 0x400A31: main (hw3.c:39)
==7879==
==7879== Invalid read of size 1
==7879==    at 0x51ADAAC: _IO_file_xsputn@@GLIBC_2.2.5 (fileops.c:1342)
==7879==    by 0x517FA6C: vfprintf (vfprintf.c:1635)
==7879==    by 0x5188C98: printf (printf.c:34)
==7879==    by 0x400BBD: main (hw3.c:72)
==7879==  Address 0x5657a36 is 6 bytes inside a block of size 30 free'd

1 个答案:

答案 0 :(得分:2)

为什么在树中插入节点后free city new成员while (!feof(fp))成员?它属于结构,当你解除分配树时它将被释放两次。

您的代码中还有其他问题:

  • for(;;)总是错的。在您的情况下,只需使用fgets()永远运行,并在new在文件末尾失败时正常退出循环。
  • 您不应该使用node作为标识符,它会混淆代码着色器(和读取器),因为它是C ++中的关键字。只需将其称为npnew->city[strlen(tmp) + 1] = '\0';
  • strcpy()最多无用,可能会导致缓冲区溢出。
  • 确实,我们不知道您的数组大小的值,但在调用strdup()来复制行块之前,您不会检查大小。这可能会调用未定义的行为。
  • 您应该使用fgets()为线段分配正确的大小,并通过一个简单的步骤复制内容。
  • 同样,在成功strtok()之后分配节点可以简化错误案例。
  • 您不检查NULL的返回值。无效的文件内容可能导致返回值为Node *importTree(const char *filename) { char line[MAXLINELENGTH]; Node *root = NULL; FILE *fp = fopen(filename, "r"); if (!fp) { printf("Error opening file.\n"); return NULL; } while (fgets(line, MAXLINELENGTH, fp) != NULL) { Node *node = malloc(sizeof(Node)); if (!node) { printf("Failed to allocate memory. Ending read.\n"); fclose(fp); exit(1); } new->left = NULL; new->right = NULL; char *tmp = strtok(line, ","); if (!tmp) { printf("Invalid file contents. Ending read.\n"); fclose(fp); exit(1); } new->zipCode = atoi(tmp); tmp = strtok(NULL, ","); if (!tmp) { printf("Invalid file contents. Ending read.\n"); fclose(fp); exit(1); } node->city = strdup(tmp); if (!(new->city)) { printf("Failed to allocate memory. Ending read.\n"); fclose(fp); exit(1); } tmp = strtok(NULL, ","); if (!tmp) { printf("Invalid file contents. Ending read.\n"); fclose(fp); exit(1); } tmp[2] = '\0'; node->state = strdup(tmp); if (!node->state) { printf("Failed to allocate memory. Ending read.\n"); fclose(fp); exit(1); } root = addNode(root, node); if (!root) { printf("Root of tree is still NULL! Ending read.\n"); fclose(fp); exit(1); } } if (!feof(fp)) { printf("File reading ended prematurely. Check for errors in the file.\n"); fclose(fp); exit(1); } fclose(fp); return root; } ,从而调用未定义的行为而不是正常的致命错误。

这是一个更简单的更正版本:

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