I am trying to get the user to re enter the the number again if they enter a value of less than ten
i am certain the problem lies with the while statement.
this what i have
printf_s("Enter the player first name: ");
scanf_s("%s", names[i], 25); //enters name and creates a newline <enter key>//
printf_s("Minimum number to stop in a turn: ");
scanf_s("%d", &min_number, sizeof(int));
do {
printf_s("please enter a number greater or egual to 10");
} while (min_number <= 10);
scanf_s("%d ", &min_number);
printf_s("please enter a number greater or equal to 10\n\n");
is a do while loop the best option or should i look at using another type of loop
enter image description here
As I was corrected, thoughfully scanf returns the number of receiving arguments successfully assigned.
Here is a fixed version:
void main(){
int num = 0;
while(num <=10){
scanf("%d ", &num);
printf_s("please enter a number greater or equal to 10\n\n");
}
}
After the authour edited the code in his question and upon his request:
do {
printf_s("please enter a number greater or egual to 10\n\n");
scanf_s("%d", &min_number);
} while (min_number < 10);
When doing the do/while loop the body of the loop is in the
do{body goes here}while(some condition);.
After another edit, on author request:
printf_s("Minimum number to stop in a turn: ");
scanf_s("%d", &min_number, sizeof(int));
while(min_number < 10){
printf_s("please enter a number greater or egual to 10\n\n");
scanf_s("%d", &min_number);
}
This will do the following it will ask for "Minimum number to stop in a turn: " if it is greater or equal to 10 it will continue after the loop. If it is not it will give clarification "please enter a number greater or egual to 10" and will receive another input. It will loop until it receive greater or egual to 10.
Related
I need to write a program where users can input their numbers as much as many as they defined, then the program will try to find which one is the lowest value and the highest value. The problems I face are:
When the program executed, the second line will wait on user's input (number) before the printf
The error "system" seems unreliable, sometimes works, sometimes doesn't work
The program only checks the last number entry, therefore it only shows the last number in min and max
You may give hints or corrections along the answers. Thank you very much.
#include <stdio.h>
float max(float num1){
float a=0, b;
if(num1 > a){
a=num1;
}
return a;
}
float min(float num2){
float x=100, y;
if(num2 < x ){
x=num2;
}
return num2;
}
int main(){
int times, interval;
float mini, maxi, data_Input;
printf("How many number would you like to type in ? : ");
scanf("%d\n",×);
printf("Type in the number: ");
scanf("%f", &data_Input);
for(interval=2; interval<=times; interval++){
printf("\nType in the number: ");
scanf("%f",&data_Input);
while(data_Input<0){
printf("Invalid Input! Please re-enter the number:");
scanf("%f",&data_Input);
}
while(data_Input>100){
printf("Invalid Input! Please re-enter the number:");
scanf("%f",&data_Input);
}
}
maxi= max(data_Input);
mini= min(data_Input);
printf("The Lowest Number is %.2f\n", mini);
printf("The Highest Number is %.2f\n", maxi);
return 0;
}
Output:
How many number would you like to type in? : 5
70
Type in the number :
Type in the number : 90.7
Type in the number : 99
Type in the number : 30
Type in the number : 50
The Lowest Number is 50.00
The Highest Number is 50.00
Okay, the thing is that you are not updating the data_input after every successive number is inputted. What you are doing is, comparing the last number to 0 or 100, which is logically incorrect.
How about you take the first number as input, then after every successive input, compare it with the min and max value. Here is the sample code.
#include <stdio.h>
float max(float num1, float num2){
if(num1 > num2){
return num1;
}
return num2;
}
float min(float num1, float num2){
if(num1 < num2){
return num1;
}
return num2;
}
int main(){
int times, interval;
float mini, maxi, data_Input;
printf("How many number would you like to type in ? : ");
scanf("%d\n",×);
printf("Type in the number: ");
scanf("%f", &data_Input);
// the first number will be minimum and maximum
mini = data_Input;
maxi = data_Input;
for(interval=2; interval<=times; interval++){
printf("\nType in the number: ");
scanf("%f",&data_Input);
// make it a composite if condition
while(data_Input<0 || data_Input>100){
printf("Invalid Input! Please re-enter the number:");
scanf("%f",&data_Input);
}
maxi= max(maxi, data_Input);
mini= min(mini, data_Input);
}
printf("The Lowest Number is %.2f\n", mini);
printf("The Highest Number is %.2f\n", maxi);
return 0;
}
The program checks the last number because you are calling the min and max function out of the for bracelets so instead you can call them inside the for bracelets like:
for(interval=2; interval<=times; interval++){
printf("\nType in the number: ");
scanf("%f",&data_Input);
while(data_Input<0){
printf("Invalid Input! Please re-enter the number:");
scanf("%f",&data_Input);
}
while(data_Input>100){
printf("Invalid Input! Please re-enter the number:");
scanf("%f",&data_Input);
}
maxi= max(data_Input);
mini= min(data_Input);
}
and instead of rewriting the same code you can just ask for the numbers inside the for loop and to initialize your interval to 1 so your main will look like:
int main(){
int times, interval;
float mini, maxi, data_Input;
printf("How many number would you like to type in ? : ");
scanf("%d\n",×);
for(interval=1; interval<=times; interval++){
printf("\nType in the number: ");
scanf("%f",&data_Input);
while(data_Input<0){
printf("Invalid Input! Please re-enter the number:");
scanf("%f",&data_Input);
}
while(data_Input>100){
printf("Invalid Input! Please re-enter the number:");
scanf("%f",&data_Input);
}
maxi= max(data_Input);
mini= min(data_Input);
}
printf("The Lowest Number is %.2f\n", mini);
printf("The Highest Number is %.2f\n", maxi);
return 0;
}
Solving the printf issue is easy. As stdout is line buffered (at least by default - it can be changed) it is flushed whenever a newline is inserted in the buffer. So, just add a newline after each message print, for example
printf("How many number would you like to type in ? : \n");
and you'll be fine.
Talking about the wrong calculation of min and max, your attempt has basically two big issues:
You are not acquiring times inputs, but only one. In fact, every time you call scanf you overwrite the same variable data_Input without performing any comparison with the previous inputs
You call min() and max() function only once, after the last input. Furthermore you try to compare the argument with a local variable that has local storage and a lifetime limited to the function itself so that at the next call it will be initialized again
In order to have a variable inside a function that it is initialized only the first time you can use static keyword. But it is not I suggest you to solve the issue.
In my opinion you don't need comparison functions: you can just update maxi and mini each time you get a new input (solving both the aforementioned issues at once):
int main(){
int times, interval;
float mini, maxi, data_Input;
printf("How many number would you like to type in ? : \n");
scanf("%d",×);
printf("Type in the number: ");
scanf("%f", &data_Input);
maxi = data_Input;
mini = data_Input;
for(interval=2; interval<=times; interval++){
printf("\nType in the number: \n");
scanf("%f",&data_Input);
while(data_Input<0){
printf("Invalid Input! Please re-enter the number:\n");
scanf("%f",&data_Input);
}
while(data_Input>100){
printf("Invalid Input! Please re-enter the number:\n");
scanf("%f",&data_Input);
}
/* Check input and, in case, update max and min */
if(data_Input > maxi)
maxi = data_Input;
if(data_Input < mini)
mini = data_Input;
}
printf("The Lowest Number is %.2f\n", mini);
printf("The Highest Number is %.2f\n", maxi);
return 0;
}
The comparison is performed inside the loop, so you don't need to store the inputs into an array.
You have plenty issues in your code.
Functions min & max do not make any sense
You do not check result of the scanf and you do not know if it was successfull
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
int main()
{
int times, interval;
float mini, maxi, data_Input;
do
{
printf("\nHow many number would you like to type in ? : ");
}while(scanf(" %d\n",×) != 1);
for(interval = 0; interval < times; interval++)
{
do
{
printf("\nType in the number: ");
}while(scanf(" %f",&data_Input) != 1 && (data_Input < 0 || data_Input > 100));
printf("%f\n", data_Input);
maxi = interval == 0 ? data_Input : MAX(maxi, data_Input);
mini = interval == 0 ? data_Input : MIN(mini, data_Input);
}
printf("The Lowest Number is %.2f\n", mini);
printf("The Highest Number is %.2f\n", maxi);
return 0;
}
When the program executed, the second line will wait on user's input (number) before the printf
Drop "\n" from "%d\n". It blocks until non-white space detected after the number and is not needed.
printf("How many number would you like to type in ? : ");
// scanf("%d\n",×);
scanf("%d",×);
printf("Type in the number: ");
If output still not seen when expected, flush it. Typically printing a '\n' will flush stdout, but not certainly.
printf("How many number would you like to type in ? : ");
fflush(stdout); // add
The error "system" seems unreliable, sometimes works, sometimes doesn't work
At least this issue: sequential check vs. checking both conditions together input validity.
Rather than
while(data_Input<0){
...
}
while(data_Input>100){
...
}
Test together.
while(data_Input<0 || data_Input>100){
...
}
The program only checks the last number entry, therefore it only shows the last number in min and max
True as code only compares values once with one call to max(). That function only compares the number to 0 rather than prior values. Likewise for min().
Consider an algorithm change
// Pseudo code for max
prompt How many number would you like to type in ? : "
get times
max_value = -INF // set to minimum possible float
for each value 1 to times
prompt "Type in the number: "
get data_Input
if data_Input > max_value
max_value = data_Input
print max_value
I'm doing a practice 'do while' program where I want the user to enter three numbers and the program will print the sum on screen. After receiving the answer, the program will ask the user if he wants to enter another three numbers to get another answer, and so on. I'm fine with this.
If user enters anything other than an integer ("A,!,%") etc, I want the program to prompt the user to re-enter a number. See comments in program.
#include <stdio.h>
/*
Do - While
This program shall ask the user to enter
three numbers and print out the sum. Entering letters
or special characters will ask the user to re-enter that
number.
example:
Enter Number 1: 2
Enter Number 2: 5
Enter Number 3: 9
Answer is 16
...
Enter Number 1: 2
Enter Number 2: yum here user incorrectly enters letters
Enter Number 2: 8 re-prompted to enter number to continue
Enter Number 3: 9
Answer is 19
*/
int main(void) {
int a,b,c,ans,n;
do{
do{
printf("Enter Number 1: ");
scanf("%d", &a);
printf("\n");
}
while((a>0) || (a<0)|| (a==0));
do{
printf("Enter Number 2: ");
scanf("%d", &b);
printf("\n");
}
while((b>0) || (b<0)|| (b==0));
do{
printf("Enter Number 3: ");
scanf("%d", &c);
printf("\n");
}
while ((c>0) || (c<0)|| (c==0));
ans = a+b+c;
printf("Answer is %d\n\n", ans);
printf("Press 1 to start over, or 0 to quit...");
scanf("%d",&n);
printf("\n");
}while (n!=0);
return 0;
}
Your program contains multiple repeated sections. Since you are gathering three numbers, you should use a for loop which runs three times.
for (int i = 0; i < 3; i++) {
/* ... */
}
Inside the for loop, you're gathering the i+1-th number each time. If the user doesn't enter a valid number, you keep trying to gather it. So the do-while loop containing the printf and scanf will go inside the for loop.
for (int i = 0; i < 3; i++) {
do {
printf("Enter Number %d: ", i+1);
scanf( /* ... */ );
} while ( /* ... */ );
}
scanf returns the number of input items successfully read as an int. So the do-while loop should repeat as long as the return value of scanf is zero. We could store the return value in a variable and check the variable in the while (...);, but we can just move the scanf itself into the while (...);. We also need an array to store the three input numbers.
int n[3];
for (int i = 0; i < 3; i++) {
do {
printf("Enter Number %d: ", i+1);
} while (scanf("%d", n+i) == 0);
}
The rest of the program would loop over the array and store the sum of the elements. You would then output the sum. This approach is robust and maintainable as changing the amount of input numbers is easy and repeated or similar code sections are eliminated.
You can use fgets to get the input and use strtol() to cast the string the user input into an int. If strtol returns 0 when the user input does not start with a number or have a number in it. From there you can check if the user input is 0 and then reprompt the user until a is not 0.
*instead of scanf()*
char num1[5];
char *end;
fgets(num1, 5, stdin);
a = strtol(num1, &end, 10);
while( a = 0){
fgets....
}
I have to write a program which takes a number as characters and then find its binary, so I used this code:
screenshotofmycode
#include<stdio.h>
#include<math.h>
int main(){
int operation,basetype,k=1,k2=1,binary=0,binary2=0,remainder,remainder2,binaryzero=0000;
char num1,num2;
printf("Please enter the number of the operation you would like to perform:");
printf("\n1. Bitwise OR\n2. Bitwise NOT\n3. Bitwise COMPARE\n4. Exit");
scanf("%d",&operation);
if(operation==1){
printf("You chose Bitwise OR operation.");
printf("\nPlease enter the first number:");
scanf(" %c",&num1);//*here i tried using %s also but it doesn't work for 2
// or more digit numbers..only for numbers from 0-9*//
printf("Please specify the base(10/16):");
scanf(" %d",&basetype);
if(basetype==10){
num1=num1-'0';//*changing from char to decimal*//
if(num1==0){
printf("\nYour first number is base 2 is %04d",binaryzero);
}
else if(num1>0){
while(num1!=0){
remainder=num1%2;
num1=num1/2;
binary=remainder*k+binary;
k=k*10;
}
printf("Your first number in base 2 is:%04d",binary);
}
else{
printf("WARNING: Your number is not valid in base 10!");
printf("\nPlease enter the first number:");
scanf(" %c",&num1);
}
}
else if(basetype==16){
if(num1==0){
printf("\nYour first number in base 2 is %04d",binaryzero);
}
else if(num1>0 && num1<10 ){{
while(num1!=0){
remainder=num1%2;
num1=num1/2;
binary=remainder*k+binary;
k=k*10;}
}
printf("\nYour first number in base 2 is %04d",binary);}
else if(num1>='A' && num1<='F'){
switch(num1){//*i will continue writing cases..*//
case 'A':printf("\nYour first number in base 2 is 1010",binary);}
}
}
}
printf("\nPlease enter the second number:");
scanf("%d",&num2);
printf("Please specify the base(10/16):");
scanf("%d",&basetype);
if(basetype==10){
if(num2==0){
printf("\nYour second number is base 2 is %04d",binaryzero);
}
else if(num2>0 && num2<10){
while(num2!=0){
remainder2=num2%2;
num2=num2/2;
binary2=remainder2*k2+binary2;
k2=k2*10;
}
printf("Your second number in base 2 is:%04d",binary2);
}
else{
printf("WARNING: Your number is not valid in base 10!");
printf("\nPlease enter the second number:");
scanf("%d",&num2);
}}
return 0;
}
Unfortunately, it only works for numbers from 0 to 9, but it should work on 2 or more digit numbers.
I have to take the number as char because the user might enter the number in base 16 and I have to ask the number first then the base type. We are not allowed to use arrays.
Use atoi() function
First include string.h
then do this
char arr[3];
scanf("%s",arr);
int num=atoi(arr);
This doesn't work for your 2-digit number:
num1=num1-'0';
I am trying to create a program which takes a number and we match that number with a variable which has a specific number stored. We need to keep doing that until the user enters the correct number that matches the number we have stored in our variable:
#include <stdio.h>
int main(void) {
int i;
int j;
int num1;
int num2 = 2;
printf("Enter number");
scanf("%d", &num1);
while (num1 == 0) {
printf("Enter number");
scanf("%d", &num1);
}
while (num1 != num2) {
for(j=1;j
printf("This is not the correct number! \n");
printf("Enter number again: ");
scanf("%d", &num1);
}
if (num1 == num2) {
printf("The numbers have matched! \n");
}
}
I am confused with how do we create a loop where we don't know how many times the user will enter an incorrect number. What I want with the loop is to display
is how many times the user enters an incorrect number. Let say if they enter 3 times.
This is not the correct number 1!
This is not the correct number 2!
This is not the correct number 3!
But we don't know how many times the user will enter an incorrect number, so what condition do I put in a loop, so it counts.
You want to create a separate counter variable that keeps track of how many times you've gone through the loop. Set it to 0 before the loop, then increment on each iteration.
int count = 0;
...
while (num1 != num2) {
count++;
printf("This is not the correct number %d! \n", count);
printf("Enter number again: ");
scanf("%d", &num1);
}
Also, I'm presuming this is a typo:
for(j=1;j
sorry for the poor title structure, couldn't think of how to word it.
Anyway, I'm facing a problem when it comes to arrays and loops in C. I need to have a user input 5 letters and store them in the first array. Then I need them to input 5 numbers between 1 and 10 and store those in a second array. Then I need to print the data of the first array the amount of times based on the corresponding number entered in the second array.
For example, if the first letter entered into the first array was A and the first number entered into the second array was 4, i need to print out AAAA.
I know that I need to use a for loop in there somewhere, I'm sure that would be better than a while loop because I will know all the data being used in the loop.
Here is my code so far, it's not much but I hate to ask a question with not even some code in it.
Cheers!
#include <stdio.h>
#include <stdlib.h>
int main() {
char input1[5];
int input2[5], i;
printf("Please enter a letter to be stored in the first array: \n");
scanf(" %c", &input1[0]);
printf("Please enter another letter to be stored in the first array: \n");
scanf(" %c", &input1[1]);
printf("Please enter another letter to be stored in the first array: \n");
scanf(" %c", &input1[2]);
printf("Please enter another letter to be stored in the first array: \n");
scanf(" %c", &input1[3]);
printf("Please enter a final letter to be stored in the first array: \n");
scanf(" %c", &input1[4]);
printf("The first array contains these values: %c, %c, %c, %c and %c. \n", input1[0], input1[1], input1[2], input1[3], input1[4]);
printf("Please enter five numbers between 1 and 10 to be stored in a second array: \n");
scanf("%d", &input2[0]);
printf("Please enter five numbers between 1 and 10 to be stored in a second array: \n");
scanf("%d", &input2[1]);
printf("Please enter five numbers between 1 and 10 to be stored in a second array: \n");
scanf("%d", &input2[2]);
printf("Please enter five numbers between 1 and 10 to be stored in a second array: \n");
scanf("%d", &input2[3]);
printf("Please enter five numbers between 1 and 10 to be stored in a second array: \n");
scanf("%d", &input2[4]);
printf("The second array contains these values: %d, %d, %d, %d and %d \n", input2[0], input2[1], input2[2], input2[3], input26[4]);
for(i=0;i<5;i++){
for(j=0;j<input2[i];j++){
printf("%c",input1[i]);
}
}
Code:
for(i=0; i < input1.length; i++) {
char charValue = input1[i];
int loopNumber = input2[i];
for(j=0 ;j < loopNumber; j++) {
printf("%c" + charValue);
}
}