无法正确释放另一个malloc的malloc [重复]

This question already has an answer here:

这个问题在这里已有答案:

String assignment in C 4 answers

C 4答案中的字符串赋值

Here's the snippet with issues.

这是有问题的片段。

int main()
{
    char** RESERV = (char**)malloc(sizeof(char*)*4);
    printf("%i, %i, %i, %i, %i", **RESERV, *RESERV, RESERV, &**RESERV, sizeof(char*));
    int i;
    for(i = 0; i < 4; i++)
    {
        RESERV[i] = (char*)calloc(sizeof(char),16);
        RESERV[i][15] = '\0';
    }
    for(i = 0; i < 4; i++)
        RESERV[i]="Iambananananananananananana";
    for(i = 0; i < 4; i++)
        printf("\r\n>%i<", RESERV[i]);
    for(i = 0; i < 4; i++)
    {
        printf("\r\n<%i>", (RESERV[i]));
        free(RESERV[i]);
    }
    free(RESERV);
}

This code free() is working fine in 32 bit , but somehow crashes horribly in 64 bit mode.

这段代码free()在32位工作正常,但在64位模式下以某种方式可怕地崩溃。

In my main program I've omitted freeing the members of the char** causing unexptected behavior every now and then, which I obviously do not want.

在我的主程序中,我省略了释放char **的成员,导致无时无刻的行为,我显然不想要。

I've tried playing around with addresses and pointers, even tried

我试过玩地址和指针,甚至尝试过

free(RESERV+(i*sizeof(char*))

Which failed too. Can someone clarify what I'm doing wrong?

哪个也失败了。有人能澄清我做错了什么吗?

3 个解决方案

#1

In your code

在你的代码中

 RESERV[i]="Iambananananananananananana";

creates the problem. It overwrites the memory allocated by malloc(). Thus,

造成问题。它会覆盖malloc()分配的内存。从而,

You face memory leak, because the malloc()ed pointer is lost.

您面临内存泄漏,因为malloc()ed指针丢失。

You cannot call free() with the changed pointer. It invokes undefined behaviour.

您无法使用更改的指针调用free()。它调用未定义的行为。

Solution:

In C, you don't assign strings, instead, you can use strcpy() to get your work done.

在C中,您不分配字符串,而是可以使用strcpy()来完成工作。

Notes:

even in case of strcpy() you cannot use "Iambananananananananananana". In this case, it will create memory overrun as destination does not have enough memory to hold it completely.

即使在strcpy()的情况下,你也不能使用“Iambanananananananananananana”。在这种情况下,它将创建内存溢出,因为目标没有足够的内存来完全保存它。
Use proper format specifiers. in your printf() statements, most of the arguments to %i are not of type int. For pointer type arguments, you should be using %p, atleast. Otherwise, it will be UB.

使用正确的格式说明符。在你的printf()语句中,%i的大多数参数都不是int类型。对于指针类型参数,您应该至少使用%p。否则,它将是UB。

#2

RESERV[i]="Iambananananananananananana";

This is not the proper way to populate this string after allocating it. You are overwriting the value of that pointer.

这不是在分配后填充此字符串的正确方法。您正在覆盖该指针的值。

strncpy(RESERV[i], "Iambananananananananananana", 15);
RESERV[i][15] = 0;

This is what you're looking for.

这就是你要找的东西。

Notice, too, that you are apparently to assign something into that pointer that is larger than the memory that you have allocated.

另请注意,您显然是在指针中指定了比您已分配的内存大的内容。

#3

The line

 RESERV[i]="Iambananananananananananana";

is incorrect - You need to do

是不正确的 - 你需要这样做

 strncpy(RESERV[i], "Iambananananananananananana", 15);
 RESERVp[i][15] = 0;

#1