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V600. The 'Foo' pointer is always not equal to NULL. Consider inspecting the condition.
The analyzer has detected a comparison of an array address to null. This comparison is meaningless and might signal an error in the program.
Consider the following code sample.
void Foo()
{
short T_IND[8][13];
...
if (T_IND[1][j]==0 && T_IND[5]!=0)
T_buf[top[0]]= top[1]*T_IND[6][j];
...
}The program handles a two-dimensional array. The code is difficult to read, so the error is not visible at first sight. But the analyzer will warn you that the "T_IND[5]!=0" comparison is meaningless. The pointer "T_IND[5]" is always not equal to zero.
After studying the V600 warnings you may find errors which are usually caused by misprints. For instance, it may turn out that the code above should be written in the following way:
if (T_IND[1][j]==0 && T_IND[5][j]!=0)The V600 warning doesn't always indicate a real error. Careless refactoring is often the reason for generating the V600 warning. Let's examine the most common case. This is how the code looked at first:
int *p = (int *)malloc(sizeof(int) *ARRAY_SIZE);
...
if (!p)
return false;
...
free(p);Then it underwent some changes. It appeared that the ARRAY_SIZE value was small and the array was able to be created on the stack. As a result, we have the following code:
int p[ARRAY_SIZE];
...
if (!p)
return false;
...The V600 warning is generated here. But the code is correct. It simply turns out that the "if (!p)" check has become meaningless and can be removed.
This diagnostic is classified as:
You can look at examples of errors detected by the V600 diagnostic. |
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