Read and print the string in C

Question

So I have been trying to read input string and then print it by not using neither scanf() nor printf() but getchar() and putchar() instead. It seems like the program is stuck in the loop, I'm not able to spot an error.

#include <stdio.h>

void getstring(char *c)
{
char inS[100];
char n;
int i = 0;  
while ((n = getchar()) != '\n') {
    inS[i] = n;
    i++;        
}
inS[i] = '\0';
i = 0;
while (inS[i] != '\0') {
    putchar(inS[i]);
    i++;
}
}
main()
{
char *prompt;
prompt = "Enter a sentence: \n";
getstring(&prompt);
printf("%s", prompt);

}

Answer 1

Maybe you are not adding \n in your stdin? Executing your code was successful. Also you are not modifying passed char *c, why? And to modify a pointer you should pass a pointer to a pointer (How do I modify a pointer that has been passed into a function in C?)

Answer 2

Your code has some problems. Here is the corrected code:

#include <stdio.h>

void getstring(void) /* You don't use the passed argument. So, I've removed it */
{
    char inS[100];
    /* `char n;` getchar() returns an int, not a char */
    int n;
    int i = 0;  
    while (i < 99 && (n = getchar()) != '\n' && n != EOF)  /* Added condition for preventing a buffer overrun and also for EOF */
    {
        inS[i] = n;
        i++;        
    }
    inS[i] = '\0';
    putchar('\n'); /* For seperating input and output in the console */
    i = 0;
    while (inS[i] != '\0')
    {
        putchar(inS[i]);
        i++;
    }
    putchar('\n'); /* For cleanliness and flushing of stdout */
}
int main(void) /* Standard signature of main */
{
    char *prompt;
    prompt = "Enter a sentence: \n";
    printf("%s", prompt); /* Interchanged these two lines */
    getstring(); /* Nothing to pass to the function */
    return 0; /* End main with a return code of 0 */
}

Note: The loop for input will end when either

• A \n(Enter) has been encountered.
• An EOF has been encountered.
• 99 characters were read from stdin.

Answer 3

First you can make all the elements of the array to NULL.

 char inS[100] = {"\0",};

You can use gets() to get the string from the console. some thing as below.

your code

while ((n = getchar()) != '\n') {
    inS[i] = n;
    i++;        
}

Modify it as:

gets(inS);

Then you can do the other operation what ever you want.

Answer 4

getchar replaces \n character with \0 whenever it sees a newline character. So n will never be \n. You should try getc or fgetc.

Answer 5

You are not printing a new line. After putchar(ch) you should use putchar('\n') to print a new line

Answer 6

#include <stdio.h>
void getstring(char *c)
{
char n;
int i = 0;
while ((n = getchar()) != '\n')
{
 *c=n;
  c++;
}
c = '\0';
main()
{
char inS[100];
char *prompt=inS;
getstring(prompt);
printf("%s", prompt);
}

strong text

In main() function only pointer is declared but it is not assigned to any symbol in other words pointer declared but it does not refer to any memory location. That problem is solved in this solution