未初始化的值是由堆栈分配创建的 [英] Uninitialised value was created by a stack allocation
问题描述
==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)
被引用的行:
int limit = atoi(argv[1]);
我不知道如何修复它.我尝试在 stackoverflow 和 google 上搜索,但找不到解决方案.
I am not sure how to fix it. I have tried searching on stackoverflow and google but I could not find the solution.
代码(来自修订历史):
The code (from revision history):
#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;
}
推荐答案
你检查过 argc
和 argv[1]
的内容了吗?argv[1]
是否保证非NULL
以适合作为atoi
的输入?由于 argv[1]
是非数字的,atoi
是否可能返回表示未初始化值的陷阱表示?
Have you checked argc
and the contents of argv[1]
? Is argv[1]
guaranteed to be non-NULL
in order to be suitable as input for atoi
? Is it possible that atoi
might be returning a trap representation representing an uninitialised value, due to argv[1]
being non-numeric?
edit:看到完整的代码后,我意识到这不是问题,你的诊断是错误的.你的问题在这里: for (i = 0; i <= count; i++) { sum += numbers[i];}
当i == count
时,numbers[i]
未初始化.这是因为 count 在上一个循环中最后一次赋值给 numbers[count]
之后递增:numbers[count] = i;计数++;
.因此,打印 sum 会导致您的消息,因为 sum 本身取决于未初始化的值.也许你的意思是 for (i = 0; i
edit: After seeing the complete code, I've realised that that's not the problem, and your diagnosis is incorrect. Your problem is here: for (i = 0; i <= count; i++) { sum += numbers[i]; }
Wheni == count
, numbers[i]
is uninitialised. This is because count is incremented after the last assignment to numbers[count]
in the previous loop: numbers[count] = i; count++;
. Hence, printing sum results in your message because sum itself depends upon an uninitialised value. Perhaps you meant for (i = 0; i < count; i++) { sum += numbers[i]; }
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