上使用malloc分配内存的sizeof（） [英] Using sizeof() on malloc'd memory

问题描述

可能重复：结果
  有关的malloc和sizeof

我想读的字符串到一个程序。当我注意到弦有时被损坏，我尝试以下code：

I am trying to read strings into a program. When I noticed that the strings were sometimes being corrupted, I tried the following code:

 void *mallocated = malloc(100);
 printf("sizeof(mallocated) = %d\n", sizeof(mallocated));

据我的程序， mallocated 的大小为 8 ，即使我分配的100个字节它。正因为如此，每当我尝试存储字符串长度超过8个字节，第8字节之后一切都会有时会消失。为什么会出现这种情况，而哪能prevent呢？

According to my program, the size of mallocated was 8, even though I allocated 100 bytes for it. Because of this, whenever I try to store a string longer than 8 bytes, everything after the 8th byte will sometimes disappear. Why is this happening, and how can I prevent it?

推荐答案

由于字符串指针的大小为8字节。下面是使用带有其相应的大小的sizeof（）的一些例子。术语size_of（）有时欺骗了不习惯使用它的人。你的情况，指针的大小为8字节。下面是一个典型的32位系统上重新presentation。

Because the size of the "string" pointer is 8 bytes. Here are some examples of using sizeof() with their appropriate "size". The term size_of() is sometimes deceiving for people not used to using it. In your case, the size of the pointer is 8 bytes.. below is a representation on a typical 32-bit system.

sizeof (char)   = 1
sizeof (double) = 8
sizeof (float)  = 4
sizeof (int)    = 4
sizeof (long)   = 4
sizeof (long long)  = 8
sizeof (short)  = 2
sizeof (void *) = 4

sizeof (clock_t)    = 4
sizeof (pid_t)  = 4
sizeof (size_t) = 4
sizeof (ssize_t)    = 4
sizeof (time_t) = 4

来源

您离开了你是如何确定你的字符串消失（字符数组）。它可能是被传递给函数，你需要通过明确的长度作为一个变量或跟踪它的地方。使用的sizeof（）不会告诉你这一点。

You are leaving out how you are determining your string is disappearing (char array). It is probably being passed to a function, which you need to pass the explicit length as a variable or track it somewhere. Using sizeof() won't tell you this.

请参阅我的<一个href=\"http://stackoverflow.com/questions/2051462/question-about-why-string-is-truncated-on-first-instance-of-0\">$p$pvious这个问题，你会看到连我缺乏有了初步的了解。

See my previous question about this and you'll see even my lack of initial understanding.

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