Free char pointer in c
Solution 1
You can use char *command;
, but then, you must allocate some memory for command
to refer to with a call to malloc()
and when you are done ith that memory, it has to be freed again with a call to free()
.
As you can see, that is a lot more work than using a fixed-size array (as you do now), but it can be made a lot safer as well, because you could create a buffer of exactly the right size, instead of hoping that the total length of the command won't exceed 100 characters.
Aside from that, your code has a problem: The filetype
pointer that the function returns points to a location within the array file_details
, but that array will be cleaned up by the compiler when executing the return
statement, so the pointer that gets returned by the function refers to some memory that is marked as "free to be used for other purposes".
If it is not a problem that the result of get_file_type
is only valid for one file at a time, you can declare the file_details
array as static
, so that it will be preserved across calls to the function.
Solution 2
When you declare an array:
char command[100];
the compiler allocates the memory for it (100 chars in this case) and command
points to the start of that memory. You can access the memory you've allocated:
command[0] = 'a'; // OK
command[99] = 'A'; // OK
command[100] = 'Z'; // Error: out of bounds
but you cannot change the value of command
:
command = NULL; // Compile-time error
The memory will be automatically freed when command
goes out of scope.
When you declare a pointer:
char *commandptr;
you only create a single variable for pointing to char
s, but it doesn't point to anything yet. Trying to use it without initialising it is an error:
commandptr[0] = 'A'; // Undefined behaviour; probably a segfault
You need to allocate the memory yourself using malloc
:
commandptr = malloc(100);
if (commandptr) {
// Always check that the return value of malloc() is not NULL
commandptr[0] = 'A'; // Now you can use the allocated memory
}
and free it when you've finished with it:
free(commandptr);
Solution 3
Why would you change it? For temporary buffers, people usually declare the arrays with [] so they don't have to worry about garbage disposal.
Comments
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Sandeep Manne almost 2 years
I am trying to find out filetypes using c code, here is the code
char *get_file_type(char *path, char *filename) { FILE *fp; char command[100]; char file_details[100]; char *filetype; sprintf(command, "file -i %s%s", path, filename); fp = popen(command, "r"); if (fp == NULL) { printf("Failed to run command\n" ); exit(1); } while (fgets(file_details, sizeof(file_details)-1, fp) != NULL) { filetype = (strtok(strstr(file_details, " "), ";")); } pclose(fp); return filetype; }
here instead of declaring command[], can I use *command? I tried to use it, but it throwed an exception. we dont need to free up variables declared like command[]? if yes how?
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Sandeep Manne over 13 yearsbut can you explain me how to do it?
-
Bart van Ingen Schenau over 13 yearsYou can, but then you must remember to call
free
in the calling function(s) to avoid a memory leak. -
jamesdlin over 13 yearsFor one thing, it probably would make sense to dynamically allocate a buffer based on the lengths of
path
andfilename
rather than silently allowing a buffer overflow. (C99 variable length arrays would solve this, though. Usingsnprintf
instead ofsprintf
would be advisable too.) -
KAction almost 12 yearsActually, stack array do not get freed, because it do not need to. To be totally honest, malloc and free are syscalls(well, wrappers to them), while stack array are not.