How to stop the loop after one try? - c

Question

I am facing some problem here. Whenever I run this code it asks me about name and date of birth. When I give it, it asks me again until I press Ctrl+D. I want this process to be stop after one time and instead it asks me, do you want to continue? If I say yes then it repeat the loop and again ask me one more time. How to do this? I tried but failed.
int main(void){
const char *listing[] = {"Name", "Date of birth"};
char data[2][51];
int done = 0;
FILE *fp;
fp = fopen("/home/bilal/Documents/file.txt", "w+");
while (!done){
for (int i = 0; i < 2; i++){
printf("Enter your %s: ", listing[i]);
if (scanf(" %50[^\n]", data[i]) != 1){
done = 1;
break;
}
}
if (!done){
fprintf(fp, "%s %s\n", data[0], data[1]);
}
}
fclose(fp);
return 0;
}

Answer 1

If I'm interpreting your post correctly, after the first prompt for name and [then] DOB, you want to inject a "Shall I continue?" question.
Here's the refactored code:
#include <stdio.h>
int
domore(void)
{
char buf[51];
int moreflg = -1;
while (moreflg < 0) {
printf("Shall I continue (y/n)? ");
if (scanf(" %50[^\n]", buf) != 1)
break;
switch (buf[0]) {
case 'y':
case 'Y':
moreflg = 1;
break;
case 'n':
case 'N':
moreflg = 0;
break;
}
}
if (moreflg < 0)
moreflg = 0;
return moreflg;
}
int
main(void)
{
const char *listing[] = { "Name", "Date of birth" };
char data[2][51];
int done = 0;
int again = 0;
FILE *fp;
#if 0
fp = fopen("/home/bilal/Documents/file.txt", "w+");
#else
fp = fopen("/tmp/file.txt", "w+");
#endif
while (!done) {
if (again) {
if (! domore())
break;
}
again = 1;
for (int i = 0; i < 2; i++) {
printf("Enter your %s: ", listing[i]);
if (scanf(" %50[^\n]", data[i]) != 1) {
done = 1;
break;
}
}
if (!done) {
fprintf(fp, "%s %s\n", data[0], data[1]);
}
}
fclose(fp);
return 0;
}

Answer 2

scanf() returns the count of parameters successfully matched and formatted, or EOF (-1) in case of an invalid input.
So, the expressions,
char data[1][50]={0};
int val=0;
int count = scanf(" %50[^\n]", data[0]); // 1 parameter to scan and format
will return 1, for a single formatted item.
To get more than 1, you have to scan in, and successfully format more than one item:
count = scanf(" %50[^\n] %d", data[0], &val); //2 parameters to scan and format.
(The return value here will be 2 if successful.)
So, if you change your loop exit criteria to match these expectations, your loop will exit properly.
if (scanf(" %50[^\n]", data[i]) == 1){//now 'done' will be set to true (1)
^^
Edit: Illustrates concepts described above:
(Note, uncomment your file functions, I did not need them for this illustration.)
Program loops until both 'name' and 'date of birth' are captured into a 2D character array. Loop exit criteria uses count of successful, single parameter scanf calls:
int main(void){
const char *listing[] = {"Name", "Date of birth"};
char data[2][51];
int done = 0;
int count = 0;//added to allow visibility of return value of scanf function.
FILE *fp;
//fp = fopen("/home/bilal/Documents/file.txt", "w+");
while (done != 2){
for (int i = 0; i < 2; i++){
printf("Enter your %s: ", listing[i]);
count = scanf(" %50[^\n]", data[i]);
if (count != EOF){
done += count;
// 'break' is not needed here, let the loop leave naturally
}
}
}
if (done == 2)//moved outside of loop because both variables
{ //need to be populated before printing
;//fprintf(fp, "%s %s\n", data[0], data[1]);
}
//fclose(fp);
return 0;
}

Answer 3

The scanf function returns the number of input items success‐fully matched and assigned. So when you type something from the keyboard, the return value of this function is always 1 in your case (you try to input the string, then every character from the keyboard is added to your string).
You can use fgets then compare with the string that content only the enter character "\n" when you want to quit the loop.
printf("Enter your %s: ", listing[i]);
if (strcmp(fgets(data[i], 51, stdin), "\n") == 0){
done = 1;
break;
}
With this code, if you want to quit the while loop, you just have to hit only ENTER.