I'm not able to understand the error

Well, put ; before if

And it will compile

And it will compile

I did it mam and it compiled I also had done mistake in variable so I changed them to convenient ones but 1 why was it showing error pointing out the precision 2 now that it has successfully compiled It is showing wrong answer

#include<stdio.h> #include<stdlib.h> int main() { int w; float ab,c; //printf("write your withdraw amount"); scanf("%d\n", &w); //printf("write your bank balance"); scanf("%.2f\n", &ab); //.2f is for points of precision c=w+0.5; if (ab>c && w%5==0) { printf("%.2f\n", ((ab-w)-0.5) ) ; //new balance on successful transaction } else{ printf("%.2f\n", ab) ; } return 0; }

I did it mam and it compiled I also had done mistake in variable so I changed them to convenient ones but 1 why was it showing error pointing out the precision 2 now that it has successfully compiled It is showing wrong answer

You normally do not put precision in a scanf

I did it mam and it compiled I also had done mistake in variable so I changed them to convenient ones but 1 why was it showing error pointing out the precision 2 now that it has successfully compiled It is showing wrong answer

What would be the right answer and what's the program's answer?

You normally do not put precision in a scanf

If for example user provides an input with precision upto 5 digits and I want to perform an operation based on only two digit decision then what to do in that case

If for example user provides an input with precision upto 5 digits and I want to perform an operation based on only two digit decision then what to do in that case

You do it normally and show as a 2 digit precision.

What would be the right answer and what's the program's answer?

#include<stdio.h> int main() { int w; float t; scanf("%d%f",&w,&t); if(w+0.5>t) { printf("%f",t); } else if(w%5!=0) { printf("%f",t); } else { printf("%f",t-w-0.5); } return 0 ; }

This answer has been accepted by the program given by some other user

#include<stdio.h> int main() { int w; float t; scanf("%d%f",&w,&t); if(w+0.5>t) { printf("%f",t); } else if(w%5!=0) { printf("%f",t); } else { printf("%f",t-w-0.5); } return 0 ; }

But does not has the precision which has been asked in the question

In the printfs you put %.2f

Just not in the scanf

Just not in the scanf

Tyvm sir my answer got accepted

Tyvm sir my answer got accepted

Yw ^.~

i'm having issue to Concatenate the strings

Hey gays .l am begging to learn C , and i need tiny ide to compile code (*.exe) . thanks

Hey gays .l am begging to learn C , and i need tiny ide to compile code (*.exe) . thanks

Code::blocks Visual Studio There are others

Put your code in pastebin.

Code::blocks Visual Studio There are others

The link 🔗 of blocks plz

The link 🔗 of blocks plz

#googleit

#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> int main() { int i = 4; double d = 4.0; char s[] = "HackerRank "; int j; float k; char l[100]; scanf("%d%f", &j,&k); for (int m=0;m<100;m++) scanf("%c\n", &l[i]); printf("%d\n", i+j); // Print the sum of the double variables on a new line. printf("%.1f\n", d+k); // Concatenate and print the String variables on a new line // The 's' variable above should be printed first. printf("%c", s); for(int o=0;o<100;o++) printf("%c", l[o]); return 0; }

Task Complete the code in the editor below. The variables , , and are already declared and initialized for you. You must: Declare variables: one of type int, one of type double, and one of type String. Read lines of input from stdin (according to the sequence given in the Input Format section below) and initialize your variables. Use the operator to perform the following operations: Print the sum of plus your int variable on a new line. Print the sum of plus your double variable to a scale of one decimal place on a new line. Concatenate with the string you read as input and print the result on a new line. Note: If you are using a language that doesn't support using for string concatenation (e.g.: C), you can just print one variable immediately following the other on the same line. The string provided in your editor must be printed first, immediately followed by the string you read as input. Input Format The first line contains an integer that you must sum with . The second line contains a double that you must sum with . The third line contains a string that you must concatenate with . Output Format Print the sum of both integers on the first line, the sum of both doubles (scaled to decimal place) on the second line, and then the two concatenated strings on the third line.

Sample Input 12 4.0 is the best place to learn and practice coding! Sample Output 16 8.0 HackerRank is the best place to learn and practice coding!

#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> int main() { int i = 4; double d = 4.0; char s[] = "HackerRank "; int j; float k; char l[100]; scanf("%d%f", &j,&k); for (int m=0;m<100;m++) scanf("%c\n", &l[i]); printf("%d\n", i+j); // Print the sum of the double variables on a new line. printf("%.1f\n", d+k); // Concatenate and print the String variables on a new line // The 's' variable above should be printed first. printf("%c", s); for(int o=0;o<100;o++) printf("%c", l[o]); return 0; }

You're printing wrong. Do it with %s and you don't need the 2nd loop

printf("%s", s); for(int o=0;o<100;o++) printf("%c", l[o]);

printf("%s", s); for(int o=0;o<100;o++) printf("%c", l[o]);

Both of them.

sir if i remove this for loop how to print the second string to be input by the stdin

With %s

printf("%s", s); printf("%s", l); return 0;

Your Output (stdout) HackerRank Expected Output HackerRank is the best place to learn and practice coding!

sir can u pls tell me how to correct this i have to attach user input of character string

scanf("%s", l); printf("%s", s); printf("%s", l); return 0; Your Output (stdout) HackerRank is expected HackerRank is the best place to learn and practice coding!

here i removed the first loop and tried to run but it is printing only 1 word not the sentence

here i removed the first loop and tried to run but it is printing only 1 word not the sentence

Okay. I'll see about that

for (int m=0;m<100;m++) { getchar(); ch=getchar(); l[i]=ch; } printf("%s", s); for (int n=0;n<100;n++) printf("%c", l[n]);

tried getchar but got this as output

pls help me out

https://pastebin.com/wFLtG0XT

why does it start a new line in this code, when i print out the data inputed in fgets

Hello world

std::cout << “Sup” << std::endl;

std::cout << “Sup” << std::endl;

That's exactly how NOT to do it!

sorry no endl

\n

sorry no endl

<<

NOT >>

cin

cin

Cin >> Cout <<

sorry

lol

all fixed now

Lol

been a while

write(1, "Heyo!/n", 6);

all fixed now

Still failing to build

https://pastebin.com/wFLtG0XT

can some help me with this

Still failing to build

Skxhbdksnsks missing main?

can some help me with this

I'm not home rn

Skxhbdksnsks missing main?

” != ";

not equal ?

not equal ?

You clearly need proper learning.

when using fgets in C, and then using printf, how do you get it to not start a new line

” != ";

Oh

True

You clearly need proper learning.

!= means not equal

!= means not equal

Yes

ok ...

Look at the quotations

when using fgets in C, and then using printf, how do you get it to not start a new line

Disappear with '/n'

Disappear with '/n'

https://pastebin.com/wFLtG0XT

Disappear with '/n'

In the string

After receiving the input you can either truncate it (awful choice. Never do this) Or you can find the character and change it to "end of the string" character ('/0').

After receiving the input you can either truncate it (awful choice. Never do this) Or you can find the character and change it to "end of the string" character ('/0').

i dont fully understand

i just updated telegram , didnt realize they have a dark mode

In the string

can you provide some syntax

Disappear with '/n'

\n

\n

did you see the pastebin

did you see the pastebin

nope, but it can't be "/n" for the new line certainly