Dynamically created C string - c

I'm trying to get an expression from the user and put it in a dynamically created string. Here's the code:
char *get_exp() {
char *exp, *tmp = NULL;
size_t size = 0;
char c;
scanf("%c", &c);
while (c != EOF && c != '\n') {
tmp = realloc(exp, ++size * sizeof char);
if (tmp == NULL)
return NULL;
exp = tmp;
exp[size-1] = c;
scanf("%c", &c);
}
tmp = realloc(exp, size+1 * sizeof char);
size++;
exp = tmp;
exp[size] = '\0';
return exp;
}
However, the first character read is a newline char every time for some reason, so the while loop exits. I'm using XCode, may that be the cause of the problem?

No, XCode is not part of your problem (it is a poor workman who blames his tools).
You've not initialized exp, which is going to cause problems.
Your code to detect EOF is completely broken; you must test the return value of scanf() to detect EOF. You'd do better using getchar() with int c:
int c;
while ((c = getchar()) != EOF && c != '\n')
{
...
}
If you feel you must use scanf(), then you need to test each call to scanf():
char c;
while (scanf("%c", &c) == 1 && c != EOF)
{
...
}
You do check the result of realloc() in the loop; that's good. You don't check the result of realloc() after the loop (and you aren't shrinking your allocation); please check every time.
You should consider using a mechanism that allocates many bytes at a time, rather than one realloc() per character read; that is expensive.
Of course, if the goal is simply to read a line, then it would be simplest to use POSIX getline(), which handles all the allocation for you. Alternatively, you can use
fgets() to read the line. You might use a fixed buffer to collect the data, and then copy that to an appropriately sized dynamically allocated buffer. You would also allow for the possibility that the line is very long, so you'd check that you'd actually got the newline.

Here on Windows XP/cc, like Michael said, it works if exp is initialized to NULL.

Here's a fixed code, with comments explaining what is different from your code in the question:
char *get_exp()
{
// keep variables with narrowest scope possible
char *exp = NULL;
size_t size = 0;
// use a "forever" loop with break in the middle, to avoid code duplication
for(;;) {
// removed sizeof char, because that is defined to be 1 in C standard
char *tmp = realloc(exp, ++size);
if (tmp == NULL) {
// in your code, you did not free already reserved memory here
free(exp); // free(NULL) is allowed (does nothing)
return NULL;
}
exp = tmp;
// Using getchar instead of scanf to get EOF,
// type int required to have both all byte values, and EOF value.
// If you do use scanf, you should also check it's return value (read doc).
int ch = getchar();
if (ch == EOF) break; // eof (or error, use feof(stdin)/ferror(stdin) to check)
if (ch == '\n') break; // end of line
exp[size - 1] = ch; // implicit cast to char
}
if (exp) {
// If we got here, for loop above did break after reallocing buffer,
// but before storing anything to the new byte.
// Your code put the terminating '\0' to 1 byte beyond end of allocation.
exp[size-1] = '\0';
}
// else exp = strdup(""); // uncomment if you want to return empty string for empty line
return exp;
}

Related

how to stop my program from skipping characters before saving them

I am making a simple program to read from a file character by character, puts them into tmp and then puts tmp in input[i]. However, the program saves a character in tmp and then saves the next character in input[i]. How do I make it not skip that first character?
I've tried to read into input[i] right away but then I wasn't able to check for EOF flag.
FILE * file = fopen("input.txt", "r");
char tmp;
char input[5];
tmp= getc(file);
input[0]= tmp;
int i=0;
while((tmp != ' ') && (tmp != '\n') && (tmp != EOF)){
tmp= getc(file);
input[i]=tmp;
length++;
i++;
}
printf("%s",input);
It's supposed to print "ADD $02", but instead it prints "DD 02".
You are doing things in the wrong order in your code: The way your code is structures, reading and storing the first char is moved out of the loop. In the loop, that char is then overwritten. In that case start with i = 1.
Perhaps you want to read the first character anyway, but I guess you want to read everything up to the first space, which might be the first character. Then do this:
#include <stdio.h>
int main(void)
{
char input[80];
int i = 0;
int c = getchar();
while (c != ' ' && c != '\n' && c != EOF) {
if (i + 1 < sizeof(input)) { // store char if the is room
input[i++] = c;
}
c = getchar();
}
input[i] = '\0'; // null-terminate input
puts(input);
return 0;
}
Things to note:
The first character is read before the loop. the loop condition and the code that stores the char then use that char. Just before the end of the loop body, the next char is read, which will then be processed in the next iteration.
You don't enforce that the char buffer input cannot be overwritten. This is dangerous, especially since your buffer is tiny.
When you construct strings char by char, you should null-terminate it by placing an explicit '\0' at the end. You have to make sure that there is space for that terminator. Nearly all system functions like puts or printf("%s", ...) expect the string to be null-terminated.
Make the result of getchar an int, so that you can distinguish between all valid character codes and the special value EOF.
The code above is useful if the first and subsequent calls to get the next item are different, for example when tokenizing a string with strtok. Here, you can also choose another approach:
while (1) { // "infinite loop"
int c = getchar(); // read a char first thing in a loop
if (c == ' ' || c == '\n' || c == EOF) break;
// explicit break when done
if (i + 1 < sizeof(input)) {
input[i++] = c;
}
}
This approach has the logic of processing the chars in the loop body only, but you must wrap it in an infinite loop and then use the explicit break.

Return a string made with a line read from input

i am trying to code a C function which returns a line read from the input as a char* . I am on Windows and i test my program in the command line by giving files as input and output of my program like this:
cl program.c
program < test_in.txt > test_out.txt
This is my (not working) function:
char* getLine(void)
{
char* result = "";
int i, c;
i = 1;
while((c = getchar()) != EOF)
{
*result++ = c;
i++;
if(c == '\n')
return result - i;
}
return result - i;
}
I was expecting it to work because i previously wrote:
char* getString(char* string)
{
//char* result = string; // the following code achieve this.
char* result = "";
int i;
for(i = 1; *result++ = *string++; i++);
return result - i;
}
And these lines of code have a correct behaviour.
Even if every answers will be appreciated, i would be really thankfull
if any of you could explain me why my getString() function works while my getLine() function doesn't.
Your function does not allocate enough space for the string being read. The variable char* result = "" defines a char pointer to a string literal ("", empty string), and you store some arbitrary number of characters into the location pointed to by result.
char* getLine(void)
{
char* result = ""; //you need space to store input
int i, c;
i = 1;
while((c = getchar()) != EOF)
{
*result++ = c; //you should check space
i++;
if(c == '\n')
return result - i; //you should null-terminate
}
return result - i; //you should null-terminate
}
You need to allocate space for your string, which is challenging because you don't know how much space you are going to need a priori. So you need to decide whether to limit how much you read (ala fgets), or dynamically reallocate space as you read more. Also, how to you indicate that you have finished input (reached EOF)?
The following alternative assumes dynamic reallocation is your chosen strategy.
char* getLine(void)
{
int ch; int size=100; size_t pos=0;
char* result = malloc(size*sizeof(char*));
while( (ch=getchar()) != EOF )
{
*result++ = ch;
if( ++pos >= size ) {
realloc(result,size+=100);
//or,realloc(result,size*=2);
if(!result) exit(1); //realloc failed
}
if( c=='\n' ) break;
}
*result = '\0'; //null-terminate
return result - pos;
}
When you are done with the string returned from the above function, please remember to free() the allocated space.
This alternative assumes you provide a buffer to store the string (and specifies the size of the buffer).
char* getLine(char* buffer, size_t size)
{
int ch;
char* result = buffer;
size_t pos=0;
while( (ch=getchar()) != EOF )
{
*result++ = ch;
if( ++pos >= size ) break; //full
if( c=='\n' ) break;
}
*result = '\0'; //null-terminate
return buffer;
}
Both avoid the subtle interaction between detecting EOF, and having enough space to store a character read. The solution is to buffer a character if you read and there is not enough room, and then inject that on a subsequent read. You will also need to null-ter
Both functions have undefined behaviour since you are modifying string literals. It just seems to work in one case. Basically, result needs to point to memory that can be legally accessed, which is not the case in either of the snippets.
On the same subject, you might find this useful: What Every C Programmer Should Know About Undefined Behavior.
Think of it this way.
When you say
char* result = "";
you are setting up a pointer 'result' to point to a 1-byte null terminated string (just the null). Since it is a local variable it will be allocated on the stack.
Then when you say
*result++ = c;
you are storing that value 'c' in to that address + 1.
So, where are you putting it?
Well, most stacks are to-down; so they grow toward lower addresses; so, you are probably writing over what is already on the stack (the return address for whatever called this, all the registers it needs restore and all sorts of important stuff).
That is why you have to be very careful with pointers.
When you expect to return a string from a function, you have two options (1) provide a string to the function with adequate space to hold the string (including the null-terminating character), or (2) dynamically allocate memory for the string within the function and return a pointer. Within your function you must also have a way to insure your are not writing beyond the end of the space available and you are leaving room for the null-terminating character. That requires passing a maximum size if you are providing the array to the function, and keeping count of the characters read.
Putting that together, you could do something similar to:
#include <stdio.h>
#define MAXC 256
char* getLine (char *s, int max)
{
int i = 0, c = 0;
char *p = s;
while (i + 1 < max && (c = getchar()) != '\n' && c != EOF) {
*p++ = c;
i++;
}
*p = 0;
return s;
}
int main (void) {
char buf[MAXC] = {0};
printf ("\ninput : ");
getLine (buf, MAXC);
printf ("output: %s\n\n", buf);
return 0;
}
Example/Output
$ ./bin/getLine
input : A quick brown fox jumps over the lazy dog.
output: A quick brown fox jumps over the lazy dog.

using getchar to receive input from file and insert into char pointer giving segmentation fault

There is a file called prog1.txt that is being read into my program. To execute, I use the name of the file and a '<' symbol to read in the file. (./a.out < prog1.txt). When i read it in however, i immediately receive a segmentation fault. Here is what I have written that is giving this issue:
char *strPtr;
while(((*strPtr = getchar()) != EOF) && (*strPtr != '\n')) {
strPtr++;
}
I have researched other questions, but I can't find a problem whose solution is usable for this. What is causing the segmentation fault??
You have a pointer char* strPtr, but it might not point to anything that you can use. You'll need to allocate some memory. char* strPtr = malloc(numChars); You should also free it after you're finished with the allocated memory: free(strPtr);
You'll need to set a maximum number of characters you can read in. In this case I use numChars.
Creating a pointer does not create something for it to point at. When using a pointer (or array syntax) it is the programmer's responsibility to ensure things are set up correctly.
getchar() returns int, and EOF is a value that cannot be represented using a char. Comparing any value of type char with EOF will therefore always fail.
You need to deal with both these concerns. You are not.
For example;
char *strPtr = malloc(10);
int length = 0;
if (strPtr != NULL)
{
int achar;
while(length < 10 && (achar = getchar()) != EOF && achar != '\n')
{
strPtr[length] = (char)achar;
++length;
}
}
/* do whatever is needed with strPtr and length*/
free(strPtr);
The check that strPtr is not NULL is to ensure the malloc() call succeeded before trying to write to the allocated array.
length is being used to ensure the code does not write to strPtr past the allocated length.
achar is used to check for EOF before converting the value to a char (not after, as in your code).
Use an array
char strPtr[10];
int temp;
int i = 0;
while( i < 10 && ((temp = getchar()) != EOF) && (temp != '\n')) {
strPtr[i] = temp;
i++;
}

Tokenizing user input in C (store in **arg)?

I'm attempting to write a simple shell like interface, that takes in a users input (by char) and stores it via a pointer to a pointer* (exactly how argv works). Here's my code:
char input[100];
char **argvInput;
char ch;
int charLoop = 0;
int wordCount = 0;
argvInput = malloc(25 * sizeof(char *));
while((ch = getc(stdin))) {
if ((ch == ' ' || ch == '\n') && charLoop != 0) {
input[charLoop] = '\0';
argvInput[wordCount] = malloc((charLoop + 1) * sizeof(char));
argvInput[wordCount] = input;
charLoop = 0;
wordCount++;
if (ch == '\n') {
break;
}
} else if (ch != ' ' && ch != '\n') {
input[charLoop] = ch;
charLoop++;
} else {
break;
}
}
If I loop through argvInput via:
int i = 0;
for (i = 0; i < wordCount; i++)
printf("Word %i: %s\n", i, argvInput[i]);
All of the values of argvInput[i] are whatever the last input assignment was. So if I type:
"happy days are coming soon", the output of the loop is:
Word 0: soon
Word 1: soon
Word 2: soon
Word 3: soon
Word 4: soon
I'm at a loss. Clearly each loop is overwriting the previous value, but I'm staring at the screen, unable to figure out why...
This line is your bane:
argvInput[wordCount] = input;
Doesn't matter that you allocate new space, if you're going to replace the pointer to it with another one (i.e. input).
Rather, use strncpy to extract parts of the input into argvInput[wordCount].
argvInput[wordCount] = input; is only making the pointer of argvInput[wordCount] point to the memory of input instead of copy the content of input into the new allocated memory. You should use memcpy or strcpy to correct your program.
After the pointer assignment the memory status looks like the image below. The memory allocated by malloc((charLoop + 1) * sizeof(char));, which are the grey ones in the graph, could not be accessed by your program anymore and this will lead to some memory leak issue. Please take care of that.
I suggest printing your argvInput pointers with %p, instead of %s, to identify this problem: printf("Word %i: %p\n", i, (void *) argvInput[i]);
What do you notice about the values it prints? How does this differ from the behaviour of argv? Try printing the pointers of argv: for (size_t x = 0; x < argc; x++) { printf("Word %zu: %p\n", x, (void *) argv[x]); }
Now that you've observed the problem, explaining it might become easier.
This code allocates memory, and stores a pointer to that memory in argvInput[wordCount]: argvInput[wordCount] = malloc((charLoop + 1) * sizeof(char)); (by the way, sizeof char is always 1 in C, so you're multiplying by 1 unnecessarily).
This code replaces that pointer to allocated memory with a pointer to input: argvInput[wordCount] = input; ... Hence, all of your items contain a pointer to the same array: input, and your allocated memory leaks because you lose reference to it. Clearly, this is the problematic line; It doesn't do what you initially thought it does.
It has been suggested that you replace your malloc call with a strdup call, and remove the problematic line. I don't like this suggestion, because strdup isn't in the C standard, and so it isn't required to exist.
strncpy will work, but it's unnecessarily complex. strcpy is guaranteed to work just as well because the destination array is allocated to be large enough to store the string. Hence, I recommend replacing the problematic line with strcpy(argvInput[wordCount], input);.
Another option that hasn't been explained in detail is strtok. It seems this is best left unexplored for now, because it would require too much modification to your code.
I have a bone to pick with this code: char ch; ch = getc(stdin); is wrong. getc returns an int for a reason: Any successful character read will be returned in the form of an unsigned char value, which can't possibly be negative. If getc encounters EOF or an error, it'll return a negative value. Once you assign the return value to ch, how do you differentiate between an error and a success?
Have you given any thought as to what happens if the first character is ' '? Currently, your code would break out of the loop. This seems like a bug, if your code is to mimic common argv parsing behaviours. Adapting this code to solve your problem might be a good idea:
for (int c = getc(stdin); c >= 0; c = getc(stdin)) {
if (c == '\n') {
/* Terminate your argv array and break out of the loop */
}
else if (c != ' ') {
/* Copy c into input */
}
else if (charLoop != 0) {
/* Allocate argvInput[wordCount] and copy input into it,
* reset charLoop and increment wordCount */
}
}

How to read a line from the console in C?

What is the simplest way to read a full line in a C console program
The text entered might have a variable length and we can't make any assumption about its content.
You need dynamic memory management, and use the fgets function to read your line. However, there seems to be no way to see how many characters it read. So you use fgetc:
char * getline(void) {
char * line = malloc(100), * linep = line;
size_t lenmax = 100, len = lenmax;
int c;
if(line == NULL)
return NULL;
for(;;) {
c = fgetc(stdin);
if(c == EOF)
break;
if(--len == 0) {
len = lenmax;
char * linen = realloc(linep, lenmax *= 2);
if(linen == NULL) {
free(linep);
return NULL;
}
line = linen + (line - linep);
linep = linen;
}
if((*line++ = c) == '\n')
break;
}
*line = '\0';
return linep;
}
Note: Never use gets ! It does not do bounds checking and can overflow your buffer
If you are using the GNU C library or another POSIX-compliant library, you can use getline() and pass stdin to it for the file stream.
A very simple but unsafe implementation to read line for static allocation:
char line[1024];
scanf("%[^\n]", line);
A safer implementation, without the possibility of buffer overflow, but with the possibility of not reading the whole line, is:
char line[1024];
scanf("%1023[^\n]", line);
Not the 'difference by one' between the length specified declaring the variable and the length specified in the format string. It is a historical artefact.
So, if you were looking for command arguments, take a look at Tim's answer.
If you just want to read a line from console:
#include <stdio.h>
int main()
{
char string [256];
printf ("Insert your full address: ");
gets (string);
printf ("Your address is: %s\n",string);
return 0;
}
Yes, it is not secure, you can do buffer overrun, it does not check for end of file, it does not support encodings and a lot of other stuff.
Actually I didn't even think whether it did ANY of this stuff.
I agree I kinda screwed up :)
But...when I see a question like "How to read a line from the console in C?", I assume a person needs something simple, like gets() and not 100 lines of code like above.
Actually, I think, if you try to write those 100 lines of code in reality, you would do many more mistakes, than you would have done had you chosen gets ;)
getline runnable example
getline was mentioned on this answer but here is an example.
It is POSIX 7, allocates memory for us, and reuses the allocated buffer on a loop nicely.
Pointer newbs, read this: Why is the first argument of getline a pointer to pointer "char**" instead of "char*"?
main.c
#define _XOPEN_SOURCE 700
#include <stdio.h>
#include <stdlib.h>
int main(void) {
char *line = NULL;
size_t len = 0;
ssize_t read = 0;
while (1) {
puts("enter a line");
read = getline(&line, &len, stdin);
if (read == -1)
break;
printf("line = %s", line);
printf("line length = %zu\n", read);
puts("");
}
free(line);
return 0;
}
Compile and run:
gcc -ggdb3 -O0 -std=c99 -Wall -Wextra -pedantic -o main.out main.c
./main.out
Outcome: this shows on therminal:
enter a line
Then if you type:
asdf
and press enter, this shows up:
line = asdf
line length = 5
followed by another:
enter a line
Or from a pipe to stdin:
printf 'asdf\nqwer\n' | ./main.out
gives:
enter a line
line = asdf
line length = 5
enter a line
line = qwer
line length = 5
enter a line
Tested on Ubuntu 20.04.
glibc implementation
No POSIX? Maybe you want to look at the glibc 2.23 implementation.
It resolves to getdelim, which is a simple POSIX superset of getline with an arbitrary line terminator.
It doubles the allocated memory whenever increase is needed, and looks thread-safe.
It requires some macro expansion, but you're unlikely to do much better.
You might need to use a character by character (getc()) loop to ensure you have no buffer overflows and don't truncate the input.
As suggested, you can use getchar() to read from the console until an end-of-line or an EOF is returned, building your own buffer. Growing buffer dynamically can occur if you are unable to set a reasonable maximum line size.
You can use also use fgets as a safe way to obtain a line as a C null-terminated string:
#include <stdio.h>
char line[1024]; /* Generously large value for most situations */
char *eof;
line[0] = '\0'; /* Ensure empty line if no input delivered */
line[sizeof(line)-1] = ~'\0'; /* Ensure no false-null at end of buffer */
eof = fgets(line, sizeof(line), stdin);
If you have exhausted the console input or if the operation failed for some reason, eof == NULL is returned and the line buffer might be unchanged (which is why setting the first char to '\0' is handy).
fgets will not overfill line[] and it will ensure that there is a null after the last-accepted character on a successful return.
If end-of-line was reached, the character preceding the terminating '\0' will be a '\n'.
If there is no terminating '\n' before the ending '\0' it may be that there is more data or that the next request will report end-of-file. You'll have to do another fgets to determine which is which. (In this regard, looping with getchar() is easier.)
In the (updated) example code above, if line[sizeof(line)-1] == '\0' after successful fgets, you know that the buffer was filled completely. If that position is proceeded by a '\n' you know you were lucky. Otherwise, there is either more data or an end-of-file up ahead in stdin. (When the buffer is not filled completely, you could still be at an end-of-file and there also might not be a '\n' at the end of the current line. Since you have to scan the string to find and/or eliminate any '\n' before the end of the string (the first '\0' in the buffer), I am inclined to prefer using getchar() in the first place.)
Do what you need to do to deal with there still being more line than the amount you read as the first chunk. The examples of dynamically-growing a buffer can be made to work with either getchar or fgets. There are some tricky edge cases to watch out for (like remembering to have the next input start storing at the position of the '\0' that ended the previous input before the buffer was extended).
How to read a line from the console in C?
Building your own function, is one of the ways that would help you to achieve reading a line from console
I'm using dynamic memory allocation to allocate the required amount of memory required
When we are about to exhaust the allocated memory, we try to double the size of memory
And here I'm using a loop to scan each character of the string one by one using the getchar() function until the user enters '\n' or EOF character
finally we remove any additionally allocated memory before returning the line
//the function to read lines of variable length
char* scan_line(char *line)
{
int ch; // as getchar() returns `int`
long capacity = 0; // capacity of the buffer
long length = 0; // maintains the length of the string
char *temp = NULL; // use additional pointer to perform allocations in order to avoid memory leaks
while ( ((ch = getchar()) != '\n') && (ch != EOF) )
{
if((length + 1) >= capacity)
{
// resetting capacity
if (capacity == 0)
capacity = 2; // some initial fixed length
else
capacity *= 2; // double the size
// try reallocating the memory
if( (temp = realloc(line, capacity * sizeof(char))) == NULL ) //allocating memory
{
printf("ERROR: unsuccessful allocation");
// return line; or you can exit
exit(1);
}
line = temp;
}
line[length] = (char) ch; //type casting `int` to `char`
length++;
}
line[length + 1] = '\0'; //inserting null character at the end
// remove additionally allocated memory
if( (temp = realloc(line, (length + 1) * sizeof(char))) == NULL )
{
printf("ERROR: unsuccessful allocation");
// return line; or you can exit
exit(1);
}
line = temp;
return line;
}
Now you could read a full line this way :
char *line = NULL;
line = scan_line(line);
Here's an example program using the scan_line() function :
#include <stdio.h>
#include <stdlib.h> //for dynamic allocation functions
char* scan_line(char *line)
{
..........
}
int main(void)
{
char *a = NULL;
a = scan_line(a); //function call to scan the line
printf("%s\n",a); //printing the scanned line
free(a); //don't forget to free the malloc'd pointer
}
sample input :
Twinkle Twinkle little star.... in the sky!
sample output :
Twinkle Twinkle little star.... in the sky!
I came across the same problem some time ago, this was my solutuion, hope it helps.
/*
* Initial size of the read buffer
*/
#define DEFAULT_BUFFER 1024
/*
* Standard boolean type definition
*/
typedef enum{ false = 0, true = 1 }bool;
/*
* Flags errors in pointer returning functions
*/
bool has_err = false;
/*
* Reads the next line of text from file and returns it.
* The line must be free()d afterwards.
*
* This function will segfault on binary data.
*/
char *readLine(FILE *file){
char *buffer = NULL;
char *tmp_buf = NULL;
bool line_read = false;
int iteration = 0;
int offset = 0;
if(file == NULL){
fprintf(stderr, "readLine: NULL file pointer passed!\n");
has_err = true;
return NULL;
}
while(!line_read){
if((tmp_buf = malloc(DEFAULT_BUFFER)) == NULL){
fprintf(stderr, "readLine: Unable to allocate temporary buffer!\n");
if(buffer != NULL)
free(buffer);
has_err = true;
return NULL;
}
if(fgets(tmp_buf, DEFAULT_BUFFER, file) == NULL){
free(tmp_buf);
break;
}
if(tmp_buf[strlen(tmp_buf) - 1] == '\n') /* we have an end of line */
line_read = true;
offset = DEFAULT_BUFFER * (iteration + 1);
if((buffer = realloc(buffer, offset)) == NULL){
fprintf(stderr, "readLine: Unable to reallocate buffer!\n");
free(tmp_buf);
has_err = true;
return NULL;
}
offset = DEFAULT_BUFFER * iteration - iteration;
if(memcpy(buffer + offset, tmp_buf, DEFAULT_BUFFER) == NULL){
fprintf(stderr, "readLine: Cannot copy to buffer\n");
free(tmp_buf);
if(buffer != NULL)
free(buffer);
has_err = true;
return NULL;
}
free(tmp_buf);
iteration++;
}
return buffer;
}
There is a simple regex like syntax that can be used inside scanf to take whole line as input
scanf("%[^\n]%*c", str);
^\n tells to take input until newline doesn't get encountered. Then, with %*c, it reads newline character and here used * indicates that this newline character is discarded.
Sample code
#include <stdio.h>
int main()
{
char S[101];
scanf("%[^\n]%*c", S);
printf("%s", S);
return 0;
}
On BSD systems and Android you can also use fgetln:
#include <stdio.h>
char *
fgetln(FILE *stream, size_t *len);
Like so:
size_t line_len;
const char *line = fgetln(stdin, &line_len);
The line is not null terminated and contains \n (or whatever your platform is using) in the end. It becomes invalid after the next I/O operation on stream.
Something like this:
unsigned int getConsoleInput(char **pStrBfr) //pass in pointer to char pointer, returns size of buffer
{
char * strbfr;
int c;
unsigned int i;
i = 0;
strbfr = (char*)malloc(sizeof(char));
if(strbfr==NULL) goto error;
while( (c = getchar()) != '\n' && c != EOF )
{
strbfr[i] = (char)c;
i++;
strbfr = (void*)realloc((void*)strbfr,sizeof(char)*(i+1));
//on realloc error, NULL is returned but original buffer is unchanged
//NOTE: the buffer WILL NOT be NULL terminated since last
//chracter came from console
if(strbfr==NULL) goto error;
}
strbfr[i] = '\0';
*pStrBfr = strbfr; //successfully returns pointer to NULL terminated buffer
return i + 1;
error:
*pStrBfr = strbfr;
return i + 1;
}
The best and simplest way to read a line from a console is using the getchar() function, whereby you will store one character at a time in an array.
{
char message[N]; /* character array for the message, you can always change the character length */
int i = 0; /* loop counter */
printf( "Enter a message: " );
message[i] = getchar(); /* get the first character */
while( message[i] != '\n' ){
message[++i] = getchar(); /* gets the next character */
}
printf( "Entered message is:" );
for( i = 0; i < N; i++ )
printf( "%c", message[i] );
return ( 0 );
}
Here is a minimal implementation to do it, the nice thing is that it will not keep the '\n', however you have to give it a size to read for security:
#include <stdio.h>
#include <errno.h>
int sc_gets(char *buf, int n)
{
int count = 0;
char c;
if (__glibc_unlikely(n <= 0))
return -1;
while (--n && (c = fgetc(stdin)) != '\n')
buf[count++] = c;
buf[count] = '\0';
return (count != 0 || errno != EAGAIN) ? count : -1;
}
Test with:
#define BUFF_SIZE 10
int main (void) {
char buff[BUFF_SIZE];
sc_gets(buff, sizeof(buff));
printf ("%s\n", buff);
return 0;
}
NB: You are limited to INT_MAX to find your line return, which is more than enough.

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