Using sizeof() on malloc'd memory
c, malloc, sizeof
Solution
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
Source
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.
See my previous question about this and you'll see even my lack of initial understanding.
Problem
Possible Duplicate: newbie questions about malloc and sizeof 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)); ``` 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?