C语言 未初始化的值是由堆栈分配创建的
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原文地址: http://stackoverflow.com/questions/15049033/
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Uninitialised value was created by a stack allocation
提问by Eduardo Bautista
==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)
The line being referenced:
被引用的行：
int limit = atoi(argv[1]);
I am not sure how to fix it. I have tried searching on stackoverflow and google but I could not find the solution.
我不知道如何解决它。我尝试在 stackoverflow 和 google 上搜索，但找不到解决方案。
The code (from revision history):
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;
}
回答by autistic
Have you checked argcand the contents of argv[1]? Is argv[1]guaranteed to be non-NULLin order to be suitable as input for atoi? Is it possible that atoimight be returning a trap representation representing an uninitialised value, due to argv[1]being non-numeric?
你检查过argc和 的内容argv[1]吗？是否argv[1]保证是非NULL以适合作为输入atoi？atoi由于argv[1]非数字，是否有可能返回表示未初始化值的陷阱表示？
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]; }
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。您的问题在这里：for (i = 0; i <= count; i++) { sum += numbers[i]; }When i == count,numbers[i]未初始化。这是因为 countnumbers[count]在上一个循环中最后一次赋值后递增：numbers[count] = i; count++;。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 < count; i++) { sum += numbers[i]; }
回答by Rohit
int limit = atoi(argv[1]) Declare it first with init value and then use it.
int limit = atoi(argv[1]) 先用init值声明然后再使用。
