未初始化的值由堆栈分配创建

Question

==13890== Conditional jump or move depends on uninitialised value(s)
==13890==    at 0x4E7E4F1: vfprintf (vfprintf.c:1629)
==13890==    by 0x4E878D8: printf (printf.c:35)
==13890==    by 0x400729: main (001.c:30)
==13890==  Uninitialised value was created by a stack allocation
==13890==    at 0x400617: main (001.c:11)

引用的行:

int limit = atoi(argv[1]);

我不知道如何解决它.我试过在stackoverflow和谷歌搜索,但我找不到解决方案.

代码(来自修订历史):

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

int main(int argc, char *argv[])
{
    if (argc != 2) {
        printf("You must pass a single integer\n");
        exit(1);
    }

    int limit = atoi(argv[1]); 
    int numbers[limit / 2];
    int count = 0;
    int i;
    for (i = 3; i < limit; i++) {
        if (i % 3 == 0 || i % 5 == 0) {
            numbers[count] = i;
            count++;
        }
    }

    int sum = 0;

    for (i = 0; i < count; i++) {
        sum += numbers[i];
    }

    printf("The sum is: %d\n", sum);

    return 0;
}

Answer 1

你有检查argc和内容argv[1]吗？是否argv[1]保证NULL不适合作为输入atoi？是否有可能atoi返回表示未初始化值的陷阱表示,因为argv[1]它是非数字的？

编辑:看到完整的代码后,我意识到这不是问题,你的诊断是不正确的.你的问题在这里:for (i = 0; i <= count; i++) { sum += numbers[i]; }什么时候i == count,numbers[i]是未初始化的.这是因为numbers[count]在上一次循环中最后一次赋值后,count会递增:numbers[count] = i; count++;.因此,打印总和会产生您的消息,因为总和本身取决于未初始化的值.也许你的意思for (i = 0; i < count; i++) { sum += numbers[i]; }