This question already has answers here:
Removing trailing newline character from fgets() input
(14 answers)
Closed 2 years ago.
#include <stdio.h>
#include <stdlib.h>
int main()
{
char name [50];
char age [50];
printf("Please enter your name:\n");
fgets(name, 20, stdin);
printf("Please enter your age:\n");
fgets(age, 20, stdin);
printf("Hello %s, you are %s years old", name, age);
return 0;
}
//sorry for such a trvial question I am new to programming, but Why does the last printf create new
lines after each placeholder?
You are getting input from stdin, so you are probably ending your entries by hitting Enter: therefore you end your line with the \n character and fgets() stores it into the name and age variables.
Related
This question already has answers here:
fgets doesn't work after scanf [duplicate]
(7 answers)
scanf() leaves the newline character in the buffer
(7 answers)
Using scanf and fgets in the same program?
(4 answers)
Closed 5 months ago.
#include <stdio.h>
int main(void) {
int numWords;
char str[100];
printf("Enter how many words:\n");
scanf("%d", &numWords);
printf("Enter %d words:", numWords);
fgets(str, 100, stdin);
return(0);
}
This program does not read fgets at all. It will print the text and scan what what the user enters into the numWords, but will not read scanf. I have tried putting a newline character after scanf but then the "Enter %d words:" will not print. I would like everything to print in order.
This is just the beginning of a more complex program, not the whole thing.
This question already has answers here:
Why does scanf ask twice for input when there's a newline at the end of the format string?
(7 answers)
Closed 1 year ago.
I have a simple code:
#include <stdio.h>
int main(void) {
int alter;
printf("Your age: ");
scanf("%d\n", &alter);
printf("your age is %d\n", alter);
}
Output is:
Your age: <entering some_number(1)> (enter)
(terminal is waiting for another value)
<entering some_number(2)> (enter)
your age is number(1)
why does it wait for me to enter 2 values?
thanks
There should be be \n in the scanf. It should be
scanf("%d", &alter);
This question already has answers here:
fgets doesn't work after scanf [duplicate]
(7 answers)
Closed 1 year ago.
CODE:
#include<stdio.h>
#include<stdlib.h>
int main(){
struct mobile{
char N[10];
int ram, pixel, price;
}B[5];
int min;
char trash;
for(int i = 0; i < 5; i++){
printf("Enter Mobile Name: ");
fgets(B[i].N, sizeof(B[i].N), stdin);
printf("Enter features (ram/camera pixels/price): ");
scanf("%d%d%d", &B[i].ram, &B[i].pixel, &B[i].price);
printf("\n");
}
}
The program is not accepting value for name of mobile second time. It prints Enter mobile name but don't take value then print Enter features and ask for value. I tried adding a second scanf above printf("\n"); but didn't work. Help please. Thanks.
Remove \n from buffer
scanf leaves a newline in the buffer, which is then read by fgets. The other problem is, that you aren't dividing the user input using a delimiter so I would put a space or a slash between the type specifiers %d:
scanf("%d/%d/%d\n", &B[i].ram, &B[i].pixel, &B[i].price);
The input should then be something like this:
Enter specs (ram/pixels/price): 8/12/500
The trailing character \n is now being read, but it isn't stored in any variable.
Remove \n from fgets() input
This doesn't cause your problem, but I would also remove the trailing \n from the fgets() input, because it's probably not supposed to be part of the phone's name.
#include <string.h>
fgets(B[i].N, sizeof(B[i].N), stdin);
B[i].N[strcspn(B[i].N, "\n")] = '\0';
This question already has answers here:
scanf() leaves the newline character in the buffer
(7 answers)
Closed 1 year ago.
This C code is not waiting for student name. It directly prints total student number. But when I comment out the first printf-scanf statement(or enter number of students:), then code is waiting for the user to enter student name.
#include <stdio.h>
int main()
{
char name[10];
int count;
printf("ENTER NUMBER OF STUDENTS:\n");
scanf("%d", &count);
printf("ENTER STUDENT NAME:\n");
scanf("%[^\n]%*c", &name);
printf("Total_Students: %d\n", count);
printf("NAME: %s\n", name);
return (0);
}
The second scanf is skipped because the newline character is being interpreted from the first scanf.
For instance, if you entered 2 for number of student, what is being entered is 2\n. The first scanf reads the number 2 and leaves the \n in the buffer which is being interpreted by the 2nd scanf.
You can simply add a space in the second scanf to get past this issue
scanf(" %[^\n]%*c", &name);
change the second scanf() argument to %s and it directly makes the last entry as NULL
.
This question already has answers here:
scanf() leaves the newline character in the buffer
(7 answers)
Closed 4 years ago.
After printing "Second text", fgets expect from me to enter a string but a program is always being stopped. That happens when I try to enter a char by scanf or getchar. What's happening?
#include <stdio.h>
int main()
{
char c[100],cc;
int x;
printf("First text\n");
scanf("%d",&x);
printf("Second text\n");
fgets(c,100,stdin);
//scanf("%c",&cc);
//cc=getchar();
printf("\n %s %d",c,x);
}
You probably press "enter" after having entered a number; scanf will then read the number, but will leave a '\n' (i.e. the newline represening "enter") in the buffer; This will be treated as an "empty" line then by gets. (BTW: use fgets instead of gets).
To overcome this enter the number and the text seperated by a space in a single line (i.e. without a newline in between).