[英]I have an unexpected buffer overrun warning, why do I have that?
I have to create a function that sum components of a vector this way:我必须以这种方式创建一个对向量的分量求和的函数:
original vector has size "size".原始向量的大小为“大小”。 I have to create a result vector dynamically with size "size/2", (because doing sum this way I've halved the original vector size).我必须动态创建一个大小为“size/2”的结果向量(因为以这种方式求和,我将原始向量大小减半)。
I've used two counters, i and j, "i" is counter of the original vector, and "j" is counter of vector "result".我使用了两个计数器,i 和 j,“i”是原始向量的计数器,“j”是向量“结果”的计数器。 I think the problem is here, because I have a buffer overrun warning.我认为问题就在这里,因为我有一个缓冲区溢出警告。
this is my code:这是我的代码:
#include <stdint.h>
#include <stdlib.h>
uint32_t* add_twobytwo(uint32_t* vect, size_t size) {
if (vect == NULL) {
return NULL;
}
uint32_t* result = malloc((size / 2) * sizeof(uint32_t));
if (result == NULL) {
return NULL;
}
size_t j = 0;
for (size_t i = 0; i < size; i += 2) {
result[j] = vect[i] + vect[i + 1];
j++;
}
return result;
}
int main(void)
{
size_t n = 6;
uint32_t* v = malloc(n * sizeof(uint32_t));
if (v == NULL) {
return NULL;
}
v[0] = 3; v[1] = 87; v[2] = 5; v[3] = 7; v[4] = 12; v[5] = 9;
uint32_t* sum = add_twobytwo(v, n);
free(v);
free(sum);
return 0;
}
green squiggle is located here:绿色曲线位于此处:
for (size_t i = 0; i < size; i += 2) {
result[j] = vect[i] + vect[i + 1];
j++;
}
I've tried to interpret the warning, and it seems that there isn't enough space in result[], but it's working properly and it does its job correctly (I've debugged line-by-line to state this).我试图解释警告,结果 [] 中似乎没有足够的空间,但它工作正常并且它正确地完成了它的工作(我已经逐行调试来说明这一点)。
You get a warning, because if size
were odd, then you would be reading elements past the end of vect
.您会收到警告,因为如果size
是奇数,那么您将在vect
末尾阅读元素。 Imagine what would happen if size was 3:想象一下如果 size 为 3 会发生什么:
i=0,j=0;
一开始,你有i=0,j=0;
. .result[0] = vect[0]+vect[1];
j++
. j++
。 j is now 1. j 现在是 1。i+=2;
. . i is now 2.我现在是 2。result[1] = vect[2]+vect[3];
However, because vect
has a size of 3, trying to read vect[3]
(which you are), will (most likely) produce a segmentation fault.但是,由于vect
的大小为 3,因此尝试读取vect[3]
(您是)将(很可能)产生分段错误。
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