英文:
Use of realloc correct approach
问题
I have the following line of code, which snippet correctly uses realloc?
我有以下行代码，哪个片段正确使用了realloc？
I have a suspicioun that the realloc call in *delete_choices line is incorrectly used is this correct?
我怀疑*delete_choices行中的realloc调用是否被错误使用了，这是正确的吗？
delete_choices = (char)malloc(files_max);
delete_choices = (char)malloc(files_max);
if (files_size >= files_max)
{
if (files_size >= files_max)
{
files_max = 2;
files = (struct dirent)realloc(files, sizeof(struct dirent) * files_max);
delete_choices = (char)realloc(delete_choices, files_max);
}
}
Is this the correct usage?
这是正确的用法吗？
delete_choices = (char)realloc(*delete_choices, files_max);
delete_choices = (char)realloc(*delete_choices, files_max);
Really confused about this
对此感到非常困惑
英文:
I have the following line of code, which snippet correctly uses realloc?
I have a suspicioun that the realloc call in *delete_choices line is incorrectly used is this correct?
*delete_choices = (char*)malloc(files_max);
if (files_size >= files_max)
{
files_max *= 2;
files = (struct dirent*)realloc(files, sizeof(struct dirent) * files_max);
*delete_choices = (char*)realloc(delete_choices, files_max);
}
Is this the correct usage?
*delete_choices = (char*)realloc(*delete_choices, files_max);
Really confused about this
答案1
得分: 3
delete_choices = (char)malloc(files_max);
if (files_size >= files_max)
{
files_max = 2;
files = (struct dirent)realloc(files, sizeof (struct dirent) * files_max);
delete_choices = (char)realloc(delete_choices, files_max);
}
正确的表达是：
```c
*delete_choices = realloc(*delete_choices, files_max);
此外，上述代码片段存在潜在的未定义行为风险，因为如果realloc()失败并返回NULL指针，您将失去对通过malloc()分配的原始内存的访问权（假设malloc()成功），从而导致内存泄漏。
解决方案：
使用临时指针来存储realloc()的返回值：
/* 是的，用法是正确的。 */
char *const tmp = realloc(*delete_choices, files_max);
if (!tmp) {
perror("realloc()");
complain();
}
*delete_choices = tmp;
注意：不要对malloc()及其相关函数的返回值进行强制类型转换。这些函数返回一个通用的void *，会自动转换为任何其他指针类型。强制类型转换是多余的，只会使代码变得混乱。
英文:
*delete_choices = (char*)malloc(files_max);
if (files_size >= files_max)
{
files_max *= 2;
files = (struct dirent*)realloc(files, sizeof (struct dirent) * files_max);
*delete_choices = (char*)realloc(delete_choices, files_max);
}
The correct expression is:
*delete_choices = realloc (*delete_choices, files_max);
Furthermore, the above code snippet risks invoking undefined behaviour because if realloc() fails and returns a NULL pointer, you'd lose access to the original memory allocated through malloc() (assuming malloc() had succeeded), and leak memory.
Solution:
Use a temporary pointer to store the return value of realloc():
/* Yes, the usage is correct. */
char *const tmp = realloc(*delete_choices, files_max);
if (!tmp) {
perror("realloc()");
complain();
}
*delete_choices = tmp;
Note: Do not cast the return of malloc() and family. These functions return a generic void * that is implicitly converted to any other pointer type. The cast is redundant and only serves to clutter one's code.
答案2
得分: 1
*delete_choices = (char*)realloc( *delete_choices, files_max ); 是正确的用法。
你传递的类型必须等于返回的类型。
在你的情况下，delete_choices 似乎是类型为 char**，因此 *delete_choices 是类型为 char*，所以这两种类型不相等。
一般情况下：t = (T)realloc( t, n );
英文:
*delete_choices = (char*)realloc( *delete_choices, files_max ); is the correct usage.
The type that you pass must be equal to the type returned.
In your case, delete_choices appears to be of type char**, therefore *delete_choices is of type char*, so these two types are not equal.
In general: t = (T)realloc( t, n );
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