I have a buffer which I receive through a serial port. When I receive a certain character, I know a full line has arrived, and I want to print it with printf method. But each line has a different length value, and when I just go with:
printf("%s", buffer);
I'm printing the line plus additional chars belonging to the former line (if it was longer than the current one).
I read here that it is possible, at least in C++, to tell how much chars you want to read given a %s, but it has no examples and I don't know how to do it in C. Any help?
I think I have three solutions:
printing char by char with a for loop
using the termination character
or using .*
QUESTION IS: Which one is faster? Because I'm working on a microchip PIC and I want it to happen as fast as possible
You can either add a null character after your termination character, and your printf will work, or you can add a '.*' in your printf statement and provide the length
printf("%.*s",len,buf);
In C++ you would probably use the std::string and the std::cout instead, like this:
std::cout << std::string(buf,len);
If all you want is the fastest speed and no formatting -- then use
fwrite(buf,1,len,stdout);
The string you have is not null-terminated, so, printf (and any other C string function) cannot determine its length, thus it will continue to write the characters it finds there until it stumbles upon a null character that happens to be there.
To solve your problem you can either:
use fwrite over stdout:
fwrite(buffer, buffer_length, 1, stdout);
This works because fwrite is not thought for printing just strings, but any kind of data, so it doesn't look for a terminating null character, but accepts the length of the data to be written as a parameter;
null-terminate your buffer manually before printing:
buffer[buffer_length]=0;
printf("%s", buffer); /* or, slightly more efficient: fputs(buffer, stdout); */
This could be a better idea if you have to do any other string processing over buffer, that will now be null-terminated and so manageable by normal C string processing functions.
Once you've identified the end of the line, you must append a '\0' character to the end of the buffer before sending it to printf.
You can put a NUL (0x0) in the buffer after receiving the last character.
buffer[i] = 0;
Related
i have tried to use k = getchar() but it doesn't work too;
here is my code
#include<stdio.h>
int main()
{
float height;
float k=0;
do
{
printf("please type a value..\n");
scanf("%f",&height);
k=height;
}while(k<0);// i assume letters and non positive numbers are below zero.
//so i want the loop to continue until one types a +ve float.
printf("%f",k);
return 0;
}
i want a if a user types letters or negative numbers or characters he/she should be prompted to type the value again until he types a positive number
Like Govind Parmar already suggested, it is better/easier to use fgets() to read a full line of input, rather than use scanf() et al. for human-interactive input.
The underlying reason is that the interactive standard input is line-buffered by default (and changing that is nontrivial). So, when the user starts typing their input, it is not immediately provided to your program; only when the user presses Enter.
If we do read each line of input using fgets(), we can then scan and convert it using sscanf(), which works much like scanf()/fscanf() do, except that sscanf() works on string input, rather than an input stream.
Here is a practical example:
#include <stdlib.h>
#include <stdio.h>
#define MAX_LINE_LEN 100
int main(void)
{
char buffer[MAX_LINE_LEN + 1];
char *line, dummy;
double value;
while (1) {
printf("Please type a number, or Q to exit:\n");
fflush(stdout);
line = fgets(buffer, sizeof buffer, stdin);
if (!line) {
printf("No more input; exiting.\n");
break;
}
if (sscanf(line, " %lf %c", &value, &dummy) == 1) {
printf("You typed %.6f\n", value);
continue;
}
if (line[0] == 'q' || line[0] == 'Q') {
printf("Thank you; now quitting.\n");
break;
}
printf("Sorry, I couldn't parse that.\n");
}
return EXIT_SUCCESS;
}
The fflush(stdout); is not necessary, but it does no harm either. It basically ensures that everything we have printf()'d or written to stdout, will be flushed to the file or device; in this case, that it will be displayed in the terminal. (It is not necessary here, because standard output is also line buffered by default, so the \n in the printf pattern, printing a newline, also causes the flush.
I do like to sprinkle those fflush() calls, wherever I need to remember that at this point, it is important for all output to be actually flushed to output, and not cached by the C library. In this case, we definitely want the prompt to be visible to the user before we start waiting for their input!
(But, again, because that printf("...\n"); before it ends with a newline, \n, and we haven't changed the standard output buffering, the fflush(stdout); is not needed there.)
The line = fgets(buffer, sizeof buffer, stdin); line contains several important details:
We defined the macro MAX_LINE_LEN earlier on, because fgets() can only read a line as long as the buffer it is given, and will return the rest of that line in following calls.
(You can check if the line read ended with a newline: if it does not, then either it was the final line in an input file that does not have a newline at the end of the last line, or the line was longer than the buffer you have, so you only received the initial part, with the rest of the line still waiting for you in the buffer.)
The +1 in char buffer[MAX_LINE_LEN + 1]; is because strings in C are terminated by a nul char, '\0', at end. So, if we have a buffer of 19 characters, it can hold a string with at most 18 characters.
Note that NUL, or nul with one ell, is the name of the ASCII character with code 0, '\0', and is the end-of-string marker character.
NULL (or sometimes nil), however, is a pointer to the zero address, and in C99 and later is the same as (void *)0. It is the sentinel and error value we use, when we want to set a pointer to a recognizable error/unused/nothing value, instead of pointing to actual data.
sizeof buffer is the number of chars, total (including the end-of-string nul char), used by the variable buffer.
In this case, we could have used MAX_LINE_LEN + 1 instead (the second parameter to fgets() being the number of characters in the buffer given to it, including the reservation for the end-of-string char).
The reason I used sizeof buffer here, is because it is so useful. (Do remember that if buffer was a pointer and not an array, it would evaluate to the size of a pointer; not the amount of data available where that pointer points to. If you use pointers, you will need to track the amount of memory available there yourself, usually in a separate variable. That is just how C works.)
And also because it is important that sizeof is not a function, but an operator: it does not evaluate its argument, it only considers the size (of the type) of the argument. This means that if you do something silly like sizeof (i++), you'll find that i is not incremented, and that it yields the exact same value as sizeof i. Again, this is because sizeof is an operator, not a function, and it just returns the size of its argument.
fgets() returns a pointer to the line it stored in the buffer, or NULL if an error occurred.
This is also why I named the pointer line, and the storage array buffer. They describe my intent as a programmer. (That is very important when writing comments, by the way: do not describe what the code does, because we can read the code; but do describe your intent as to what the code should do, because only the programmer knows that, but it is important to know that intent if one tries to understand, modify, or fix the code.)
The scanf() family of functions returns the number of successful conversions. To detect input where the proper numeric value was followed by garbage, say 1.0 x, I asked sscanf() to ignore any whitespace after the number (whitespace means tabs, spaces, and newlines; '\t', '\n', '\v', '\f', '\r', and ' ' for the default C locale using ASCII character set), and try to convert a single additional character, dummy.
Now, if the line does contain anything besides whitespace after the number, sscanf() will store the first character of that anything in dummy, and return 2. However, because I only want lines that only contain the number and no dummy characters, I expect a return value of 1.
To detect the q or Q (but only as the first character on the line), we simply examine the first character in line, line[0].
If we included <string.h>, we could use e.g. if (strchr(line, 'q') || strchr(line, 'Q')) to see if there is a q or Q anywhere in the line supplied. The strchr(string, char) returns a pointer to the first occurrence of char in string, or NULL if none; and all pointers but NULL are considered logically true. (That is, we could equivalently write if (strchr(line, 'q') != NULL || strchr(line, 'Q') != NULL).)
Another function we could use declared in <string.h> is strstr(). It works like strchr(), but the second parameter is a string. For example, (strstr(line, "exit")) is only true if line has exit in it somewhere. (It could be brexit or exitology, though; it is just a simple substring search.)
In a loop, continue skips the rest of the loop body, and starts the next iteration of the loop body from the beginning.
In a loop, break skips the rest of the loop body, and continues execution after the loop.
EXIT_SUCCESS and EXIT_FAILURE are the standard exit status codes <stdlib.h> defines. Most prefer using 0 for EXIT_SUCCESS (because that is what it is in most operating systems), but I think spelling the success/failure out like that makes it easier to read the code.
I wouldn't use scanf-family functions for reading from stdin in general.
fgets is better since it takes input as a string whose length you specify, avoiding buffer overflows, which you can later parse into the desired type (if any). For the case of float values, strtof works.
However, if the specification for your deliverable or homework assignment requires the use of scanf with %f as the format specifier, what you can do is check its return value, which will contain a count of the number of format specifiers in the format string that were successfully scanned:
§ 7.21.6.2:
The [scanf] function returns the value of the macro EOF if an input failure occurs
before the first conversion (if any) has completed. Otherwise, the function returns the
number of input items assigned, which can be fewer than provided for, or even zero, in
the event of an early matching failure.
From there, you can diagnose whether the input is valid or not. Also, when scanf fails, stdin is not cleared and subsequent calls to scanf (i.e. in a loop) will continue to see whatever is in there. This question has some information about dealing with that.
How to scan total line from user input with c program?
I tried scanf("%99[^\n]",st), but it is not working when I scan something before this scan statment.It worked if this is the first scan statement.
How to scan total line from user input with c program?
There are many ways to read a line of input, and your usage of the word scan suggests you're already focused on the scanf() function for the job. This is unfortunate, because, although you can (to some extent) achieve what you want with scanf(), it's definitely not the best tool for reading a line.
As already stated in the comments, your scanf() format string will stop at a newline, so the next scanf() will first find that newline and it can't match [^\n] (which means anything except newline). As a newline is just another whitespace character, adding a blank in front of your conversion will silently eat it up ;)
But now for the better solution: Assuming you only want to use standard C functions, there's already one function for exactly the job of reading a line: fgets(). The following code snippet should explain its usage:
char line[1024];
char *str = fgets(line, 1024, stdin); // read from the standard input
if (!str)
{
// couldn't read input for some reason, handle error here
exit(1); // <- for example
}
// fgets includes the newline character that ends the line, but if the line
// is longer than 1022 characters, it will stop early here (it will never
// write more bytes than the second parameter you pass). Often you don't
// want that newline character, and the following line overwrites it with
// 0 (which is "end of string") **only** if it was there:
line[strcspn(line, "\n")] = 0;
Note that you might want to check for the newline character with strchr() instead, so you actually know whether you have the whole line or maybe your input buffer was to small. In the latter case, you might want to call fgets() again.
How to scan total line from user input with c program?
scanf("%99[^\n]",st) reads a line, almost.
With the C Standard Library a line is
A text stream is an ordered sequence of characters composed into lines, each line consisting of zero or more characters plus a terminating new-line character. Whether the last line requires a terminating new-line character is implementation-defined. C11dr §7.21.2 2
scanf("%99[^\n]",st) fails to read the end of the line, the '\n'.
That is why on the 2nd call, the '\n' remains in stdin to be read and scanf("%99[^\n]",st) will not read it.
There are ways to use scanf("%99[^\n]",st);, or a variation of it as a step in reading user input, yet they suffer from 1) Not handling a blank line "\n" correctly 2) Missing rare input errors 3) Long line issues and other nuances.
The preferred portable solution is to use fgets(). Loop example:
#define LINE_MAX_LENGTH 200
char buf[LINE_MAX_LENGTH + 1 + 1]; // +1 for long lines detection, +1 for \0
while (fgets(buf, sizeof buf, stdin)) {
size_t eol = strcspn(buf, "\n"); **
buf[eol] = '\0'; // trim potential \n
if (eol >= LINE_MAX_LENGTH) {
// IMO, user input exceeding a sane generous threshold is a potential hack
fprintf(stderr, "Line too long\n");
// TBD : Handle excessive long line
}
// Use `buf[[]`
}
Many platforms support getline() to read a line.
Short-comings: Non C-standard and allow a hacker to overwhelm system resources with insanely long lines.
In C, there is not a great solution. What is best depends on the various coding goals.
** I prefer size_t eol = strcspn(buf, "\n\r"); to read lines in a *nix environment that may end with "\r\n".
scanf() should never be used for user input. The best way to get input from the user is with fgets().
Read more: http://sekrit.de/webdocs/c/beginners-guide-away-from-scanf.html
char str[1024];
char *alline = fgets(str, 1024, stdin);
scanf("%[^'\n']s",alline);
I think the correct solution should be like this. It is worked for me.
Hope it helps.
I dont understand why I am getting an output like this: StackOver↨< as snprintf should take care of null termination as expected output is StackOver. I am using devcpp IDE.
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main(void)
{
char buffer[10];
printf("%d\n", sizeof(buffer));
snprintf(buffer, sizeof(buffer), "%s", "StackOverflow");
printf("%s", buffer);
return 0;
}
The C Standard states that the copied string shall be nul-terminated:
7.21.6.5 The snprintf function
...
Description
The snprintf function is equivalent to fprintf , except that the
output is written into an array (specified by argument s ) rather than
to a stream. If n is zero, nothing is written, and s may be a
null pointer. Otherwise, output characters beyond the n-1st
are discarded rather than being written to the array, and a null
character is written at the end of the characters actually written
into the array. If copying takes place between objects that
overlap, the behavior is undefined.
It appears you are running with an outdated and/or buggy C runtime library as the snprintf() implementation doesn't seem to properly implement the required behavior.
This code is working fine for me. The buffer only has space for 10 characters so sprintf is only able to write the first 9 characters that you tell it to write ("StackOver"). At the tenth character it stores a terminating null character, since every C string must be null-terminated.
The only suggestion I would make is adding a newline when printing the string at the end:
printf("%s\n", buffer);
The lack of a newline at the end might be the reason why your IDE is showing you that ↨ character.
If you want the buffer to fit "StackOverflow" you need to allocate it to something larger.
So I am trying to read a defined number of characters from the input. Let's say that I want to read 30 characters and put them in to a string. I managed to do this with a for loop, and I cleaned the buffer as shown below.
for(i=0;i<30;i++){
string[i]=getchar();
}
string[30]='\0';
while(c!='\n'){
c=getchar(); // c is some defined variable type char
}
And this is working for me, but I was wondering if there is another way to do this. I was researching and some of them are using sprintf() for this problem, but I didn't understand that solution. Then I found that you can use scanf with %s. And some of them use %3s when they want to read 3 characters. I tried this myself, but this command only reads the string till the first empty space. This is the code that I used:
scanf("%30s",string);
And when I run my program with this line, if I for example write: "Today is a beatiful day. It is raining, but it's okay i like rain." I thought that the first 30 characters would be saved in to the string. But when i try to read this string with puts(string); it only shows "Today".
If I use scanf("%s",string) or gets(string) that would rewrite some parts of my memory if the number of characters on input is greater than 30.
You can use scanf("%30[^\n]",s)
Actually, this is how you can set which characters to input. Here, carat sign '^' denotes negation, ie. this will input all characters except \n. %30 asks to input 30 characters. So, there you are.
The API you're looking for is fgets(). The man page describes
char *fgets(char *s, int size, FILE *stream);
fgets() reads in at most one less than size characters from stream and stores them into the buffer pointed to by s. Reading stops after an EOF or a newline. If a newline is read, it is stored into the buffer. A terminating null byte ('\0') is stored after the last character in the buffer.
I'd like to write a wrapper function for the mvwprint/mvwchgat ncurses functions which prints the message in the specified window and then changes its attributes.
However, mvwchgat needs to know how many characters it should change - and I have no idea how to tell mvwchgat how long the formatted string is, since a strlen() on, for instance, "abc%d" obviously returns 5, because strlen doesn't know what %d stands for ...
In C99 or C11, you can use a line like this:
length = snprintf(NULL, 0, format_string, args);
From the manual of snprintf (emphasis mine):
The functions snprintf() and vsnprintf() do not write more than size bytes (including the terminating null byte ('\0')). If the output was truncated due to this limit then the return value is the number of characters (excluding the terminating null byte) which would have been written to the final string if enough space had been available. Thus, a return value of size or more means that the output was truncated.
Since we are giving snprintf 0 as the size, then the output is always truncated and the output of snprintf would be the number of characters that would have been written, which is basically the length of the string.
In C89, you don't have snprintf. A workaround is to create a temporary file, or if you are in *nix open /dev/null and write something like this:
FILE *throw_away = fopen("/dev/null", "w"); /* On windows should be "NUL" but I haven't tested */
if (throw_away)
{
fprintf(throw_away, "<format goes here>%n", <args go here>, &length);
fclose(throw_away);
} /* else, try opening a temporary file */
You can't know in advance how long your string will be:
printf("abc%d", 0); //4 chars
printf("abc%d", 111111111);//12 chars
All with the same format string.
The only sure way is to sprintf the text in question into a buffer, strlen(buffer) the result and printf("%s", buffer); the result to screen.
This solution avoids double formatting at the cost of allocating long enough buffer.