So I have declared a pointer *f of type FILE and now I say that that pointer is equal to fopen("text.txt", "r"). So since a pointer stores an address, is fopen giving back the address of a file?
FILE *f;
f = fopen("text.txt","r");
is fopen giving back the adress of a file?
No. There is no such thing as "address of a file".
What fopen returns is a pointer to dynamically allocated opaque structure FILE, which describes how to get the contents of the file. This description is opaque in a sense that it provides no useful info to you. But routines such as fgets(), fread(), etc. know how to use that info to get the actual file contents.
fclose deallocates this structure, so if you have matching fopen and fclose there are no memory leaks (from these functions).
The function fopen is returning the address of an object of type FILE. According to §7.21.1 ¶2 of the ISO C11 standard, this object must be capable of
recording all the information needed to control a stream, including its file position indicator, a pointer to its associated buffer (if any), an error indicator that records whether a read/write error has occurred, and an end-of-file indicator that records whether the end of the file has been reached;
The exact size and contents of this object may differ from compiler to compiler and is of no interest to the average programmer. All that the average programmer must know is that they must pass the pointer returned by fopen to other I/O functions provided by the C standard library.
The standard does not specify how FILE should be defined. The only thing it says is what result you will get when you pass an object of that type to various functions. This means that this type may be different in various implementations. This is ONE way that is used:
typedef struct _iobuf
{
char* _ptr;
int _cnt;
char* _base;
int _flag;
int _file;
int _charbuf;
int _bufsiz;
char* _tmpfname;
} FILE;
It comes from MinGW32.
Yes, it's giving back the address of a FILE object. Now that type is opaque. That is, you are meant to not know the actual contents of the data structure being referenced and hence must never dereference the pointer (happily, the compiler will not let you get away with this).
Related
As far as I understand passing a pointer to a function essentially passes the copy of the pointer to the function in C. I have a FILE pointer that I pass to a function func(), func() reads a line from a file and then when we return to main(). I read another line from the file using the same FILE pointer.
However, while I would imagine that I'd read the line exactly from before func() was called, I actually read the next line after what func() had read. Can you please explain why FILE pointer behaves this way?
This is my code:
#include <stdio.h>
#define STR_LEN 22
void func(FILE *fd);
int main() {
FILE *fd;
char mainString[STR_LEN];
if (!(fd = fopen("inpuFile", "r"))) {
printf("Couldn't open file\n");
fprintf(stderr, "Couldn't open file\n");
}
func(fd);
fgets(mainString, STR_LEN, fd);
printf("mainString = %s\n", mainString);
fclose(fd);
return 0;
}
void func(FILE *fd) {
char funcString[STR_LEN];
fgets(funcString,STR_LEN, fd);
printf("funcString = %s\n", funcString);
}
However, while I would imagine that I'd read the line exactly from before func was called ...
I can't imagine why you would imagine that. What if the FILE* references a network connection that has no replay capability at all where reading is consumption. Where would the line be stored such that you could read it again? There would be absolutely no place to put it.
Not only would I not imagine that, it's kind of crazy.
As far as I understand passing a pointer to a function essentially passes the copy of the pointer to the function in C.
Correct. But a copy of a pointer points to the very same object. If I point to a car and you copy me, you're pointing to the very same one and only car that I'm pointing to.
Because FILE pointer points on some data that gets changed when the file is read/written.
So the pointer doesn't change (still points to the handler structure of the file) but the data pointed by the structure does.
Try passing pointer as const FILE * you'll see that you cannot because fread operation (and others) alter the pointed data.
One way would be to duplicate the file descriptor, which dup does, but doesn't work on buffered FILE object, only raw file descriptors.
The problem is in your initial statement:
As far as I understand passing a pointer to a function essentially passes the copy of the pointer to the function in C.
This does not change much, as whatever you are accessing as a pointer, still holds the location of the FILE you are accessing, the whole point of using pointers as arguments for a function in C, is so that you can modify a certain value outside the scope of a function.
For example, common usage of an integer pointer as a function argument:
void DoSomethingCool(int *error);
Now using this code to catch the error would work like this:
int error = 0;
DoSomethingCool(&error);
if(error != 0)
printf("Something really bad happened!");
In other words, the pointer will actually modify the integer error, by accessing it's location and writing to it.
An important thing to keep in mind to avoid these kinds of misunderstandings is to recognize that all a pointer is, is essentially the address of something.
So you could (in theory, by simplifying everything a lot) think of an int * as simply an int, the value of which happens to be an address of some variable, for a FILE *, you can think of it as an int, where the value of the int is the location of the FILE variable.
FILE *fd is a pointer only in the sense that its implementation uses C construct called a "pointer". It is not a pointer in the sense of representing a file position.
FILE *fd represents a handle to a file object inside the I/O library, a struct that includes the actual position of the file. In a grossly simplified way, you can think of fd as a C pointer to a file pointer.
When you pass fd around your program, I/O routines make modifications to the file position. This position is shared among all users of fd. If a func() makes a change to that position by reading some data or by calling fseek, all users of the same fd will see the updated position.
When reading K&R, I became interested in how the file position is determined. By file position, I mean where in the file the stream is currently reading or writing. I think it must have something to do with the file pointer, or the piece of data it's pointing to. So I checked stack overflow, and find the following answer:
Does fread move the file pointer?
The answer indicates that file pointer will change with the change of file position. This makes me very confused, because in my understanding, a file pointer for a certain file should always point to the same address, where information about this file is stored. So I wrote a small piece of code, trying to find the answer:
#include<stdio.h>
int main(void)
{
char s[1000];
FILE *fp,*fp1,*fp2;
fp = fopen("input","r");
fp1 = fp; /* File poiter before fread */
fread(s,sizeof(char),100,fp);
fp2 = fp; /* File pointer after fread */
printf("%d\n",(fp1 == fp2) ? 1 : -1);
}
It gives the output 1, which I believe indicates that the file pointer actually doesn't move and is still pointing to the same address. I have also changed the fread line to be a fseek, which gave the same output. So does file pointer move with the change of file position, or where am I wrong in the verifying process?
Thanks!
I think you are confusing the general concept of pointers in C, vs. the nomenclature of a "file pointer". FILE is just a structure that contains most of the "housekeeping" attributes that the C stdio runtime library needs to interact with when using the stdio functions such as, fopen(), fread(), etc. Here is an example of the structure:
typedef struct {
char *fpos; /* Current position of file pointer (absolute address) */
void *base; /* Pointer to the base of the file */
unsigned short handle; /* File handle */
short flags; /* Flags (see FileFlags) */
short unget; /* 1-byte buffer for ungetc (b15=1 if non-empty) */
unsigned long alloc; /* Number of currently allocated bytes for the file */
unsigned short buffincrement; /* Number of bytes allocated at once */
} FILE;
Note that this may be somewhat platform-dependent, so don't take it as gospel. So when you call fopen(), the underlying library function interacts with the O/S's file system APIs and caches relevant information about the file, buffer allocation, etc, in this structure. The fopen() function allocates memory for this structure, and then returns the address of that memory back to the caller in the form of a C Pointer.
Assigning the pointers values to another pointer has no effect on the attributes inside the FILE structure. However, the FILE structure, internally, may have indexes or "pointers" to the underlying O/S file. Hence, the confusion in terminology. Hope that helps.
You are right fp is never changed by fread, fseekor other f... functions. Except, of course, if you do fp = fopen(...), but then you are assigning the return value of fopen to fp and then fp changes of course.
Remember, in C parameters are passed by value, so fread cannot change it's value.
But fread does change the internal structure fp points to.
You made some confusion between a file pointer, under common definition, and the pointer in the file.
Normally with the term file pointer we refer to a pointer to a FILE structure. That structure contains all variables necessary to manage file access. This structure is created upon a successful opening of a file, and remains the same (same address) for all the time until you fclose() the file (when became undefined).
Inside the FILE structure there are many pointers that points to the file block on disk and to the position inside the current record. These pointers, managed by file I/O routines, changes when file is accessed (read or write).
And these pointers are that to which the answer you cited refers.
Given the following short example
FILE *p = fopen("foo.txt", "r");
FILE f = *p;
int i;
fscanf(p, "%i", &i); // works just fine
fscanf(&f, "%i", &i); // segmentation fault
I have read a little about FILE, FILE * as well as the actual struct type _IO_FILE, but it isn't really clear to me what is causing the segmentation fault in the second call to fscanf.
So aside from p and &f containing different addresses and unless this is involved (which I think it is), what is the difference between &f and p in this context?
The C Standard (C99 7.19.3/6, C11 7.21.3/6) says:
The address of the FILE object used to control a stream may be significant; a copy of a FILE object need not serve in place of the original.
So you have been warned.
You can consider FILE* as an opaque pointer sensu lato, you should never try to do what you are doing.
From cppreference.com:
C streams are objects of type std::FILE that can only be accessed and manipulated through pointers of type std::FILE* (Note: while it may be possible to create a local object of type std::FILE by dereferencing and copying a valid std::FILE*, using the address of such copy in the I/O functions is undefined behavior).
I'm stuck for sometime in this code:
#include <stdio.h>
const char *fn; int a;
int exists(const char *fname)
{
FILE *file;
if (file = fopen(fname, "r"))
{
fclose(file);
return 1;
}
return 0;
}
main(){
printf("Name:\n");
scanf("%s",&fn);
a=exists(&fn)
if(a==0){
fopen(&fn,"w");
fprintf(fn,"banhdhsjha");
}
}
When I try to run the program, it works until fprintf(fn,"banhdhsjha");, but it crashes here (Windows gives me an error) and the compiler (CodeBlocks) gives me the following notice:
passing argument 1 of 'fopen' from incompatible pointer type.
I am trying to make fprintf write data in the file, but I don't know how to do it. Can you help me?
There are quite a few problems in your code.
You are passing &fn to scanf. This &fn is a pointer to pointer of const char ** type. This does not make sense. Format specifier %s requires a const char * argument, not const char **. You are basically using the pointer fn itself as a target buffer for file name (4 or 8 bytes long, depending on your platform's pointer size). Most likely you type in longer file name than that "buffer" can accommodate. That overrides memory in your program and leads to unpredictable (undefined) behavior.
The proper form is probably scanf("%s", fn), but the problem is that you never allocated memory for the target buffer. You have to make sure that fn points to a char buffer of sufficient size to hold your file name.
Your fopen(&fn,"w") suffers from the same problem (and that is what the compiler is telling you). fopen expects an argument of const char * type and you are passing a const char ** instead. The proper form is fopen(fn,"w"), but again, see 2. Also, fopen return a file handle that you are supposed to store and use later. You are ignoring (discarding) the return value of fopen. That also makes no sense. You need an additional FILE * variable to store the return of fopen. You already know that, judging by what you did in exists function, but somehow you are ignoring that knowledge in your main.
Your fprintf call also makes no sense at all. fprintf requires file handle (of FILE * type) as its first argument. Instead, you are trying to pass it a file name. That's not going to work and that's also going to trigger a diagnostic message from the compiler. You are supposed to store the return value of fopen (as I said in 3) and pass it to fprintf.
Your exists(&fn) call suffers from the same problem as 1 and 3 and produces the same diagnostic message as your fopen call.
Stop trying to write random code. Where did you get the idea to pass file name to fprintf instead of file handle? Read the documentation for each function you are trying to use and act accordingly.
I noticed that I had used a char* variable instead of a FILE* variable in my code when using fopen and fgets, but my code works. I am wondering why this is? A section of my code is as follows.
...
char* filePath = ac->filepath;
char* line = malloc(sizeof(char) * MAX_CHAR_PER_LINE) ;
filePath = fopen(filePath, "r"); // we are assigning the result to a char*, not FILE*
if (filePath == NULL) {
printf ("\n[%s:%d] - error opening file '%s'", __FILE__, __LINE__, filePath);
printf ("\n\n");
exit (1);
}
while ((fgets(line, MAX_CHAR_PER_LINE, filePath) != NULL)) {
...
Both a char* and a FILE* simply store a memory address. C has fairly weak typing (Edit: this was a misunderstanding on my part, see comments below) so it lets you assign pointers without worrying about the type they point to.
fopen returns the address of a FILE object and you store that address somewhere (in your case it is in a char*). When you use the address in fgets it still has the address of the FILE object so everything will work as expected.
Your compiler is very permissive. It should have complained that the FILE* to char* conversion is invalid.
Anyway, assuming the compiler accepts this implicit conversion, the answer is quite simple.
By the C standard, a char* and a void* pointer can hold, without loss, any pointer value.
So, you can convert from SOMETYPE* to char* and then, back from char* to SOMETYPE*, and you get the same pointer you initially had.
Actually, on most systems, all pointers are equivalent and you can freely convert from one to another.
FILE* is a small and opaque value of pointer type. Probably an actual pointer to an internal data structure, but this may be implemented differently (e.g. it might be a UNIX file descriptor converted to a pointer).
The STDIO functions just expect that you use the same opaque FILE* value you got from fopen, in fread/fwrite/fclose.
The FILE* is converted to char* when assigned to filePath.
The char* is converted to FILE* as the first parameter of fgets (again, your compiler should complain), and so, gets back to its initial value.
Suggestion: Use higher levels of errors/warnings in your compiler and fix your code.