The printf function in C is not working as intended [closed] - c

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Can someone help me with this issue? For some reason, no matter what I try, doing printf isn't printing in my code. I've been trying to use the flush method, but that seems to just cause other issues. Is there anything else I can do?
int main(void) {
char **line;
size_t bufsize=MAXBUF;
printf("Type your name: ");
getline(&line,&bufsize,stdin);
}

Answer 1

You are trying to pass in an empty pointer, this pointer isn't pointing to any meaningful address at the start of the program. So the problem with this is that even if you were successful, it would overwrite some random address in memory with the input, and this is almost certainly what you don't want as it will cause a segmentation fault or a crash.
However that's not the reason why it's not compiling. It's not compiling because you are trying to pass in char *** to a function that expects char **.
So what I would do is:
#include <stdio.h>
int main(void) {
char *line=NULL;
size_t bufsize=0;
printf("Type your name: ");
if(getline(&line,&bufsize,stdin)==-1){
puts("Error: User cancelled input.");
}
else{
printf("Entered: %s",line);
}
free(line);
return 0;
}
So what this code does is that it creates a pointer to an array of characters but it sets the pointer to point to NULL so that the program can see it's not pointing anywhere, and then its address (&line) is given to the getline function, this turns the char * type into a char ** which is required by the getline function. Whenever you put a & in front of something in C, it adds an extra * to the type.
You probably know this already, but if a function states that it returns something, you should always return something, even if it's nonsense otherwise depending on the type that should be returned, that can sometimes cause a crash.
There was nothing wrong with your use of printf, that was fine. printf doesn't necessarily fflush to stdout, often the \n character will trigger a flush, but it's implementation specific, so if your program crashes before an fflush is written to stdout, then you may never see the last printf. You can try an fflush(stdout); before the getline function, but after the getline function may not work because it may crash before then.
In your case what's happening is that before the printf writes to the screen the getline takes in the address of the pointer (or pointer to a pointer), dereferences it to a pointer, isn't able to, so before you have a chance to enter in any keystrokes it crashes, losing the stdout pipe, so you end up never seeing your prompt.
When i run the altered code, above, i get:
XXXX#dell:~$ gcc Deleteme.c
XXXX#dell:~$ ./a.out
Type your name: test1
Entered: test1
XXXX#dell:~$
And it runs correctly under Linux. The key lesson in all this is that, terminals are slow to update on the screen, so some time in the past, the decision was made to separate out the functionality of printf.
There are two components to printf:
The part that writes the text to stdout.
The part that updates the terminal with stdout (the flush).
This question is interesting because 1) succeeded, but the system crashed before 2) was completed.

Answer 2

The documentation for getline says the first parameter must be a char **, but you pass it the address of line. Since line is a char **, &line is a char ***, which is the wrong type.

Answer 3

getline() allocates and re-allocates memory. The address of the length and the buffer need to reflect that.
int main(void) {
char *line = NULL;
size_t bufsize = 0;
printf("Type your name: ");
fflush(stdout); // Flush output before reading.
ssize_t len = getline(&line, &bufsize, stdin);
if (len < 0) Handle_error_with_code();
...
free(line); // When done, free memory
}
There is no need to allocate any memory before calling getline().
getline() expects line, bufsize to reflect allocated data via malloc() and friends. It is easy enough to start with NULL, 0 here.

Answer 4

First you are passing the address of line to the getline(), the documentation says that the first argument is the address of the first character position where the input string will be stored. It’s not the base address of the buffer, but of the first character in the buffer.
This pointer type (a pointer-pointer or the ** thing) causes massive confusion.
second your line pointer does not point to anything this should fix it:
char *line = (char *)malloc(bufsize * sizeof(char));
now line is pointing to valid memory address which we requested via malloc() function.