writing a program that will be finding min, max, avg of values entered by user. Having trouble writing something that will check to make sure there are only postive integers entered and produce an error message. heres my for statement that is reading the input so far:
for (int value = 0; value <= numofvals; ++value) {
printf("Value %d: %f\n", value, val_input);
scanf("%f", &val_input);
}
mind you I've been learning code for about 3 weeks and was just introduced to loops this week so my understanding is rudimentary at best!
First, don't use scanf. If stdin doesn't match what it expects it will leave it in the buffer and just keep rereading the same wrong input. It's very frustrating to debug.
const int max_values = 10;
for (int i = 0; i <= max_values; i++) {
int value;
if( scanf("%d", &value) == 1 ) {
printf("Got %d\n", value);
}
else {
fprintf(stderr, "I don't recognize that as a number.\n");
}
}
Watch what happens when you feed it something that isn't a number. It just keeps trying to read the bad line over and over again.
$ ./test
1
Got 1
2
Got 2
3
Got 3
foo
I don't recognize that as a number.
I don't recognize that as a number.
I don't recognize that as a number.
I don't recognize that as a number.
I don't recognize that as a number.
I don't recognize that as a number.
I don't recognize that as a number.
I don't recognize that as a number.
Instead, use fgets to reliably read the whole line and sscanf to parse it. %f is for floats, decimal numbers. Use %d to recognize only integers. Then check if it's positive.
#include <stdio.h>
int main() {
const size_t max_values = 10;
int values[max_values];
char buf[1024];
size_t i = 0;
while(
// Keep reading until we have enough values.
(i < max_values) &&
// Read the line, but stop if there's no more input.
(fgets(buf, sizeof(buf), stdin) != NULL)
) {
int value;
// Parse the line as an integer.
// If it doesn't parse, tell the user and skip to the next line.
if( sscanf(buf, "%d", &value) != 1 ) {
fprintf(stderr, "I don't recognize that as a number.\n");
continue;
}
// Check if it's a positive integer.
// If it isn't, tell the user and skip to the next line.
if( value < 0 ) {
fprintf(stderr, "Only positive integers, please.\n");
continue;
}
// We got this far, it must be a positive integer!
// Assign it and increment our position in the array.
values[i] = value;
i++;
}
// Print the array.
for( i = 0; i < max_values; i++ ) {
printf("%d\n", values[i]);
}
}
Note that because the user might input bad values we can't use a simple for loop. Instead we loop until either we've read enough valid values, or there's no more input.
Something easy like this may work for you:
int n;
int ret;
for (;;) {
ret = scanf("%d", &n);
if (ret == EOF)
break;
if (ret != 1) {
puts("Not an integer");
for (;;)
if (getchar() == '\n')
break;
continue;
}
if (n < 0) {
puts("Not a positive integer");
continue;
}
printf("Correct value %d\n", n);
/* Do your min/max/avg calculation */
}
/* Print your results here */
This is just an example and assumes you do not need to read floating point numbers and then check if they are integers, as well as a few other things. But for starters, it is simple and you can work on top of it.
To break out of the loop, you need to pass EOF (typically Ctrl+D in Linux/macOS terminals, Ctrl+Z in Windows ones).
An easy and portable solution
#include <limits.h>
#include <stdio.h>
int get_positive_number() {
char buff[1024];
int value, ch;
while (1) {
printf("Enter positive number: ");
if (fgets(buff, 1023, stdin) == NULL) {
printf("Incorrect Input\n");
// Portable way to empty input buffer
while ((ch = getchar()) != '\n' && ch != EOF)
;
continue;
}
if (sscanf(buff, "%d", &value) != 1 || value < 0) {
printf("Please enter a valid input\n");
} else {
break;
}
}
return value;
}
void solution() {
// Handling malformed input
// Memory Efficient (without using array to store values)
int n;
int min = INT_MAX;
int max = INT_MIN;
double avg = 0;
printf("Enter number of elements: ");
scanf("%d", &n);
getc(stdin);
int value;
for (int i = 0; i < n; i++) {
value = get_positive_number();
if (value > 0) {
if (min > value) {
min = value;
}
if (max < value) {
max = value;
}
avg += value;
}
}
avg = avg / n;
printf("Min = %d\nMax = %d\nAverage = %lf\n", min, max, avg);
}
int main() {
solution();
return 0;
}
Output:
Enter number of elements: 3
Enter positive number: 1
Enter positive number: 2
Enter positive number: a
Please enter a valid input
Enter positive number: -1
Please enter a valid input
Enter positive number: 1
Min = 1
Max = 2
Average = 1.333333
Related
First, I apologize if the question doesn't make sense as my English isn't that good...
My question is, how do we print out different things depending on the user input?
What I'm trying to do is: when user inputs integer, the program prints out the inputted number. When the user inputs something that's not integer (like symbols and characters), the program prints out "not integer".
my current idea (pseudo-code) is as follows:
`int main(){
int value;
printf("Enter numbers");
scanf("%d", &value);
if(value is integer){
printf("%d", value);
} else {
printf("not integer");
}
return 0;
}`
what gets me is the scanf; by using %d, I'm assuming that the user will input an integer values, but the user can input values that are not integers so I can't make a comparison using the if statement if( value is integer). How can I make a comparison that will determine whether the inputted value is integer or not?
I don't know if this is a good thing or not.
You can use ASCII to check if the input type is an integer or not
(between 48 - 57 in ASCII)
it will be like this
char value;
int flag = 0; //to check true or false (0 means false, and 1 means true)
printf("Enter numbers");
scanf("%c", &value);
for(int i = 48; i <= 57; i++){
if(value == i){
flag = 1;
break;
}
}
if(flag == 1){
printf("%c", value);
} else {
printf("not integer");
}
How do you print different things depending the user input?
Step 1: Read the line of user input
char buf[100];
if (fget(buf, sizeof buf, stdin)) {
// something was entered
Step 2: test the string
char *end;
long value = strtol(buf, *end);
// If the end is the same as the beginning, no conversion occurred.
if (end == buf) {
puts("not integer");
}
printf("%ld\n", value);
}
}
Additional code could look for input that occurred after the integer. Also code could test for a large number that overflowed the long range.
The code is as follows. It caters for different situations like inputting negative numbers and decimal numbers:
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main() {
char input[20];
int wrongFlag = 0;
scanf("%s", input);
if (input[0] == '0' && strlen(input) > 1) {
wrongFlag = 1;
//for number starts with 0, and string length>1 eg: 010
}
for (int i = 0; i < strlen(input); i++) {
if (i == 0 && (input[i] == '-' && strlen(input) > 2 && input[i + 1] == '0')) {
//check first round only: negative number with length >2 and starts with 0 eg: -010.
wrongFlag = 1;
continue;
}
if (i != 0 && !isdigit(input[i])) {
//check following rounds, check if it is not digit
wrongFlag = 1;
break;
}
}
if (wrongFlag) {
printf("Not integer");
}
else {
printf("integer");
}
return 0;
}
Try this it works for me.
#include<stdio.h>
#include<string.h>
int main()
{
int i;
char value[50];
int len;
printf("Enter maximum 50 digits\n");
/* enter the values you wanted*/
printf("Enter the value: ");
gets(value);
len = strlen(value);
/*it will iterate upto the end of the user input*/
for(i=0;i<len;i++)
{
if(48<value[i] && value[i]<=57)
{
if(i==(len-1))
printf("It's an integer");
}
else{
printf(" Not an integer");
break;
}
}
return 0;
}
I'm solving CS50 (problemset 1) i.e water.c. It asks user to write a program that prompts the user for the length of his or her shower in minutes (as a positive integer) and then prints the equivalent number of bottles of water (as an integer).
1 min of shower = 12 bottles consumed
MAIN PROBLEM: The problem is that we have to ensure that the user inputs a positive number of minutes otherwise it keeps on re-prompting his back to input/scanf statement. As long as he enters he enters length<=0, I can re-prompt him back using while(length<=0) condition but as he enters a character i.e abc123 in input my code keeps on executing. Any solutions??
>
#include <stdio.h>
int main()
{ int length=0;
int min=12;
int bottle=0;
printf("Enter length of his or her shower in minutes");
scanf("%d", &length);
while (length <= 0){
printf("Enter length of his or her shower in minutes");
scanf("%d", &length);
}
bottle= (min*length);
printf("%d", bottle);
return 0;
}
You can solve this by reading a string first, and then extracting any number:
#include <stdio.h>
int main(void)
{
int length = 0;
char input[100];
while(length <= 0) {
printf("Enter length: ");
fflush(stdout);
if(fgets(input, sizeof input, stdin) != NULL) {
if(sscanf(input, "%d", &length) != 1) {
length = 0;
}
}
}
printf("length = %d\n", length);
return 0;
}
Program session:
Enter length: 0
Enter length: -1
Enter length: abd3
Enter length: 4
length = 4
Crucially, I always check the return value from scanf, the number of items successfully converted.
If you don't care about Inputs like 1f then the Above Answers are ok For you, but if you do not want to accept this kind of Input, then the following approach does something like that:
#include<stdio.h>
int checkInput(void);
int main(void){
int number = checkInput();
printf("\nYour number is\t%d\n",number);
return 0;
}
int checkInput(void){
int option,check;
char c;
do{
printf("Please type a number:\t");
if(scanf("%d%c",&option,&c) == 0 || c != '\n'){
while((check = getchar()) != 0 && check != '\n' && check != EOF);
printf("\tI sayed a Number please\n\n");
}else{
if ( option < 1){
printf("Wrong input!\n");
}else{
break;
}
}
}while(1);
return option;
}
Output:
Please type a number: 1f
I sayed a Number please
Please type a number: f1
I sayed a Number please
Please type a number: -1
Wrong input!
Please type a number: 1
Your number is 1
You don't need the first prompt outside the loop because you have already initialised length to zero, so the loop will prompt at least once.
On most platforms other then Wndows, you need to flush stdout to show text not terminated with a newline.
scanf will return so long as a newline character is buffered and %d alone will not consume the newline, so you need to ensure that any remaining characters up to and including the newline are flushed to prevent an endless loop.
It is good practice to check the return value from scanf() since it makes no guaranteed about not modifying its arguments even when a conversion fails.
It is not clear why min is a variable here sine it is initialised but never re-assigned, but presumably that may be the case in the final program?
#include <stdio.h>
int main( void )
{
int length = 0 ;
int min = 12 ;
int bottle = 0 ;
while( length <= 0 )
{
int converted = 0 ;
printf( "Enter length of his or her shower in minutes: " ) ;
fflush( stdout ) ;
converted = scanf( "%d", &length ) ;
if( converted != 1 )
{
length = 0 ;
}
while( (c = getchar()) != '\n' && c != EOF ) { } // flush line buffer
}
bottle = min * length ;
printf( "%d", bottle ) ;
return 0;
}
int min = 0;
do {
printf("Enter minutes: ");
scanf("%i", &min);
} while(min <= 0);
//programs resumes after this.
everyone!
I hope someone can help me figure out something in C language.
This is my first seriously homework in IT, I have no experience and I'm learning in e-studies, so teacher help isn't very available.
I need to develop console application in C language. User need to input 10 integer numbers, if insert number isn't integer, need to output error and again re-enter new number until all 10 integer numbers will be inserted.
Everything works in case if I say that these 10 numbers can't be 0 (I make this to be sure that my if-else statement working), but won't work when I want that every input number will be check if it is integer or not.
How can I do it right.
Please help
so far my code look like this:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int i;
float f;
int numbers[10];
for (i = 0; i < 10; i++)
{
scanf ("%d", &numbers[i]);
if (numbers[i] != 0)
{
scanf ("*%d", &numbers[i]);
}
else
{
printf ("\nError!Entered number is't integer \n");
printf ("\nPlease insert number again \n");
scanf("%*d", &numbers[i]);
}
}
}
#include <stdio.h>
int main(void) {
int i = 0;
int val;
char ch;
int numbers[10];
while(i < 10) {
val = scanf("%d", numbers + i); // read the integer into a[i]
if(val != 1) {
while((ch = getchar()) != '\n') // discard the invalid input
; // the null statement
printf("Error! Entered number is not an integer.\n");
printf("Please enter an integer again.\n");
val = scanf("%d", numbers + i);
continue;
}
++i;
}
// process the numbers array
return 0;
}
I write this line again
val = scanf("%d", numbers + i);
Now it works how I need. Great - thanks a lot
There are several techniques you might use:
Read the number as a string and reject if it contains characters not suitable for an integer. The use sscanf() to convert the string to integer.
Read the number as a float and reject if it is out of integer range or it has a non-integer value.
Read the input character by character and build up an integer value. If invalid characters appear, reject the value.
scanf returns the number of input items successfully matched and assigned. You can check this value for 1 for each call of scanf. If the value is 0, then you should discard the input to clear the stdin buffer and read input again.
#include <stdio.h>
#include <ctype.h>
int main(void) {
int i = 0;
int val;
char ch;
int numbers[10];
while(i < 10) {
// read an integer and the first non-numeric character
val = scanf("%d%c", numbers + i, &ch);
// if the number of items assigned by scanf is not 2 or if
// the first non-numeric character is not a whitespace, then
// discard the input and call read input again.
// for example input of type 32ws are completely discarded
if(val != 2 || !isspace(ch)) {
while((ch = getchar()) != '\n') // discard the invalid input
; // the null statement
printf("Error! Entered number is not an integer.\n");
printf("Please enter an integer again.\n");
continue;
}
++i;
}
// process the numbers array
return 0;
}
Although I am not entirely clear on the details of your question, here is an outline of code similar to what you want:
int main(void)
{
int i;
int numbers[10];
int sum = 0;
for(i=0; i<10; ++i)
{
printf("Enter #%d:\n", i+1);
scanf("%d", numbers+i);
if (numbers[i] % 2 == 0) // Then Number is even
{
sum += numbers[i];
}
}
printf("The sum of only the even numbers is %d\n", sum);
getch();
return 0;
}
To read an int, suggest fgets() then sscanf() or strtol()
#include <stdio.h>
#include <stdlib.h>
int main(void) {
int i;
int numbers[10];
for (i = 0; i < 10; ) {
char buffer[50];
if (fgets(buffer, sizeof buffer, stdin) == NULL) break;
int n; // number of `char` parsed
if (sscanf(buffer, "%d %n", &numbers[i], &n) != 1 || buffer[n] != '\0') {
printf("Error! Entered number is not an integer.\n");
printf("Please enter an integer again.\n");
continue;
}
i++;
}
return 0;
}
The strtol() approach. This detects overflow issues:
if (fgets(buffer, sizeof buffer, stdin) == NULL) break;
char *endptr;
errno = 0;
long num = strtol(buffer, &endptr, 10);
if (errno || num < INT_MIN || num > INT_MAX) Handle_RangeError();
if (buffer == endptr || *endptr != '\n') Handle_SyntaxError();
numbers[i] = (int) num;
Recommend making a int GetInt(const char *prompt) function that can be used repeatedly.
User input is evil. Do not trust it until well vetted.
My aim is to accept 4-digit numbers, and 4-character strings (string should not contain digits or special characters)
If an invalid input is given the program should not terminate and it must allow the user to enter the details and continue until he wish to terminate.
I am able to find whether the input is a digit.
if(scanf("%d",&input)!=1)
{
printf("enter the number please");
... // I have option to re enter using while and flags
}
else
{
// I continue my work
...
}
To check it is four digits I have tried using the commands
i=0;
num = input;
while(num>0)
{
i = i+1;
num = num/10;
}
if(i==4){
...//I continue
}
else
printf("please enter four digit");
I have no idea of checking the same for characters. (I know how to check its length using strlen())
Please help me with the code in C. (Also help me to reduce/optimize the above logic to check whether the input is a 4-digit number)
I believe you want 2 inputs a number and a string. You can do that as
int number= 0;
char string[10] = { 0 };
do {
printf("please enter four digit");
scanf("%d", &number);
if(number >=1000 && number<= 9999)
break;
} while(1);
do {
printf("please enter four character string");
fgets(string, sizeof(string), stdin);
if(strlen(string) == 4)
break;
} while(1);
To check it is four digit number you can simply put a check whether the number lies between 1000 and 9999. (I am assuming you don't want the number to start with 0.)
strtol can help:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
char s[32], *p;
int x;
fgets(s, sizeof(s), stdin);
if ((p = strchr(s, '\n')) != NULL)
*p = '\0';
x = (int)strtol(s, &p, 10);
if ((p - s) == 4) {
printf("%d\n", x);
} else {
printf("Please enter four digit\n");
}
return 0;
}
char input[16];
int ok = 1, k = 0;
if (scanf("%s", input) > 0 && strlen(input) == 4) {
// check if it's a word
for (; k < 4; k++)
if (!isalpha(input[k])) {
// check if it's a number
for (int k = 0; k < 4; k++)
if (!isdigit(input[k]))
ok = 0;
break;
}
}
else ok = 0;
if (!ok)
printf("invalid input, please enter a 4-digit number or 4-letter word");
else {
printf("valid input");
...
}
You can use gets()1 fgets() to get the whole line and check line length. If the first character is between '0' and '9' then check the remaining if they are 3 numbers too. If the first character is a valid character in string then check the 3 remaining chars if it's also valid in string.
1See Why is the gets function so dangerous that it should not be used?
I've searched in and out of these forums but am still having trouble. My understanding of C is still very basic. I'm creating a small program that takes 3 numerical values entered by the user and then calculates the highest. I nailed that.
I now want to ensure that the user enters only integer values. I managed to get the prog to keep prompting the user to re-enter the value until it is within the specified numerical range (for example, any number between 1 and 1000 inclusive, piece of cake) but that's not good enough. I used 3 while loops to test each value, but this only works as long as the input is of type integer.
The trick is I cant use built in functions. It needs to be manual (sorry, poor choice of words) I tried to use char variables and x = getchar(); to get the ASCII value and test it in a condition but I can't get it working in a loop. (while/ do-while)
I also tried using a "for loop" and array variables but once again am struggling to get it to keep prompting the user.
I've also tried to test the value returned by scanf to see if its integer but my knowledge level of correct C syntax is level: noob. Either my loops don't loop or they loop infinitely.
Here is some sample code:
int x, y, z =0;
printf("Enter the first number:\n");
scanf("d", &x);
while (condition) /* Here is where I need to determine that the entered val is false */
{
printf("Wrong input. Re-enter a valid value.\n");
x =0;
scanf("%d", &x); /*user re-prompted */
}
I'm getting the idea that I'll have to use ASCII and a loop, but I just can't get to it. Also, the values entered get sent to a function for comparing and are then returned.
Could someone give me some advice and a few tips please?
Much thanks
You would have to use something like fgets, and strtol:
long someValue;
char *bufEnd = NULL;
char buf[128]; // max line size
do {
printf("enter a value: ");
fgets(buf, 128, stdin);
someValue = strtol(buf, &bufEnd, 10); // base 10
} while (bufEnd == buf || *bufEnd != '\n');
printf("got value: %li", someValue);
What we are doing here is we are tapping into strtol's capability to tell us where it stopped parsing, by passing in bufEnd.
Then, we are making sure that bufEnd doesn't point to the beginning of buf (in which case, it didn't start with a number), and also checking to make sure that bufEnd points to \n, or the end of the line (making sure that the user didn't enter something like 123abc, which strtol would interpret as 123). You may wish to trim buf of whitespace characters first, however.
You're absolutely on the right track with "scanf()". Just check the return value. If you don't get the expected #/values, then you got invalid input:
char found = FALSE;
int ival;
double x;
while (!found)
{
printf("Please enter a valid integer: ");
if (scanf("%d", &ival) !=1) {
printf ("Invalid! Please re-enter!\n");
continue;
}
printf("Please enter a valid floating point number: ");
if (scanf("%lf", &x) !=1) {
printf ("Invalid! Please re-enter!\n");
continue;
}
found = TRUE;
}
Here's my solution. It safe against buffer overflow and straightforward .
#include <stdio.h>
#define LEN 10
int main() {
int a;
char str[LEN];
fgets( str, LEN, stdin );
while ( !sscanf( str, "%d", &a ) )
fgets( str, 10, stdin );
printf("Num is : %d\n", a);
return 0;
}
#include <stdio.h>
#include <limits.h>
#include <ctype.h>
int getInteger(int* err){
int ch;
int n;//int32
int takeNum, sign;
long long int wk;//long long int as int64
wk=0LL;
*err = 0;
takeNum = 0;//flag
sign = 1;//minus:-1, other:1
/* //skip space character
while(EOF!=(ch=getchar()) && (ch == ' ' || ch == '\t' || ch == '\n'));
ungetc(ch, stdin);
*/
while(EOF!=(ch=getchar())){
if(ch == '-'){
if(takeNum != 0){//in input number
*err = 1;
break;
}
if(sign == -1){//already sign
*err = 2;
break;
}
sign = -1;
continue;
}
if(ch >= '0' && ch <= '9'){//isdigit(ch) in ctype.h
if(takeNum == 0)
takeNum = 1;
wk = wk * 10 + (ch - '0')*sign;
if(INT_MAX < wk || INT_MIN > wk){//overflow
*err = 3;
break;
}
continue;
}
if(ch != '\n'){//input other [-0-9]
*err = 4;
}
break;
}
if(takeNum == 0){//not input number
*err = 5;
} else {
n=wk;
}
while(ch != '\n' && EOF!=(ch=getchar()));//skip to newline
return n;
}
int getValue(const char* redoprompt, int low, int high){
int num, err=0;
while(1){
num = getInteger(&err);
if(err || low > num || high < num)
printf("%s", redoprompt);
else
break;
}
return num;
}
#define max(x,y) ((x)>(y))? (x) : (y)
int main(){
const char *error_message = "Wrong input. Re-enter a valid value.\n";
int x, y, z, max;
x = getValue(error_message, 1, 1000);
y = getValue(error_message, 1, 1000);
z = getValue(error_message, 1, 1000);
max = max(max(x,y), z);
printf("max:%d\n", max);
return 0;
}