I'm new in C programming language.
I need to get every digit separately that user have entered.
Here is my code:
#include <stdio.h>
int main()
{
int n[100];
printf("Enter a number: ");
scanf("%d",&n);
printf("%d %d %d",n[1],n[2],n[3]);
return 0;
} //i know that my code is not assigning like i want.
and now for example user entered a number like 123, i want the output like 1 2 3, How can i assign every digit to n[i] ? Without using string to int or int to string like atoi? Here is what Im going to do: User will enter a number and the program will search from Matrix 100x100 in row or column. i think i need to get the every digit separately to search.
No need to go to character array. The lats digit of a number n can be computed using n%10. Then you can remove the last digit using n /= 10. So this cycle would print the digits in reverse order:
void print_rev_digits(int n) {
while (n) {
printf("%d\n", n%10);
n /= 10;
}
}
And using a stack you can print the digits in the correct order. You can also use recursion for this(which will use stack for you). I am deliberately not posting a complete solution.
In this case you should read the user input character by character:
#include <stdlib.h>
#include <stdio.h>
#include <ctype.h>
int main()
{
char input[100];
int n[100];
printf("Enter a number: ");
if (fgets(input, sizeof(input), stdin)) { // attempt to read a line
int i;
for (i = 0; input[i]; i++) { // for each entered character
if (input[i] >= '0' && input[i] <= '9') { // is a digit
n[i] = input[i] - '0';
printf("%d ", input[i] - '0');
}
else if (isspace(input[i])) // end of entered integer
break;
else {
printf(stderr, "Input is not a number\n");
return -1;
}
}
printf("\n");
} else {
fprintf(stderr, "User did not enter valid input.\n");
}
return 0;
}
Related
I am new to C so I am having a little difficulty!
I want to take an integer input from the user and add 7 to each of the digit in the input. All of that works, but the digits are printing in the reverse order.
How do i make the digits print in the correct order? I checked other similar questions on Stack overflow but it does not seem to work. Thanks in advance!
int main(void)
{
int numToEncrypt;
printf("Please input a 4-digit number you wish to encrypt: ");
scanf("%d", &numToEncrypt);
while (numToEncrypt > 0)
{
int digit = numToEncrypt % 10;
// do something with digit
digit = (digit + 7)%10;
numToEncrypt /= 10;
printf("number is: %d \n",digit);
}
}
)
Converting the string input to an integer and back is pointless. Just work with the data as a string. eg:
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int main(void)
{
int c;
if( getenv("V") ) {
printf("Please input the number you wish to encrypt: ");
fflush(stdout);
}
while( (c = getchar()) != EOF ) {
if( isspace(c) ) {
fflush(stdout);
} else if( isdigit(c) ) {
c = '0' + (c - '0' + 7) % 10;
} else {
fprintf(stderr, "Invalid input: %c", c);
return EXIT_FAILURE;
}
putchar(c);
}
}
Note that a huge advantage of doing this is that it is easy to work with ten million digit integers. You will not be able to do that using scanf to convert the string into an integer.
One way is using a variable to specify which digit to process.
#include <stdio.h>
int main(void)
{
int numToEncrypt;
int delta = 1000; // for 4-digit number
printf("Please input a 4-digit number you wish to encrypt: ");
scanf("%d", &numToEncrypt);
while (delta > 0)
{
int digit = (numToEncrypt / delta) % 10;
// do something with digit
digit = (digit + 7)%10;
delta /= 10;
printf("number is: %d \n",digit);
}
}
As this is homework, you could use recursion:
#include <stdio.h>
void print_recursive(int num)
{
// print leading digits
if (num>9)
{
print_recursive(num/10);
}
// print last digits
printf("number is: %d\n", (num+7)%10);
}
int main(void)
{
int number;
printf("Please input a 4-digit number you wish to encrypt: ");
scanf(" %d", &number); // note: You should check the return value!
print_recursive(number);
}
It is not limited to 4 digits.
For a simple program like this, one usually does not bother with a lot of design. However, it is also beneficial to practice software design on simple problems like this, since the knowledge extends to more complicated programs. This is an application of divide and conquer (as a problem solving strategy, not the computer algorithm). The idea being that smaller problems are simpler than larger ones.
In this case, you consider encapsulating the work of "encrypting" to a function, and have the function return the encrypted value. We'll just implement a stub for now, and fill it in later.
int encryptBy7(int input) {
int output = 0;
return output;
}
In addition, we can encapsulate the work of "printing" to a function. And, this is your actual question, if we think about it critically.
void printDigitByDigit(int num, const char *msg) {
printf("stub\n");
}
So your main program would look like:
int main(void) {
int numToEncrypt;
int numEncrypted;
printf("Please input a 4-digit number you wish to encrypt: ");
scanf("%d", &numToEncrypt);
numEncrypted = encryptBy7(numToEncrypt);
printDigitByDigit(numEncrypted, "number is");
return 0;
}
So, your algorithm to encrypt seems to work, so let's just code that up in a way that it stores it as a number.
int encryptBy7(int input) {
int output = 0;
int pow10 = 1;
/* Original program didn't deal with 0 */
if (input == 0) return 0;
while (input > 0) {
int digit = input % 10;
// do something with digit
digit = (digit + 7)%10;
input /= 10;
// build output
output += digit * pow10;
pow10 *= 10;
}
return output;
}
So, now we get to the meat of your question, which is about how to print out the digits starting with the most significant digit. If we see how we built up the output in the previous function, we can reverse the same process of looking at the powers of 10 to find the most significant digit, and then work backwards from there.
void printDigitByDigit(int input, const char *msg) {
int pow10 = 1;
int x = input;
// Find the position of the most significant digit
while (x > 10) {
pow10 *= 10;
x /= 10;
}
// Divide by the input appropriate power of 10 to
// find and print the corresponding digit
while (pow10 > 0) {
int digit = (input / pow10) % 10;
printf("%s: %d\n", msg, digit);
pow10 /= 10;
}
}
Of course, you are free to choose to try to do this as a single program inside of main as you had originally attempted, and the result would probably be a shorter program. I'll leave that as an exercise. However, I would argue that breaking up the program into tasks will provide you more benefit in the long run, and itself is a problem solving tool. Each function becomes easier to think about, and thus an easier problem to solve.
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 am currently trying to finish a code where a user inputs two 5 digit long numbers. The code then checks to see if there are any identical numbers in the same spot for the two numbers and displays how many identical numbers there are in the same spot of the two inputs. (ex. comparing 56789 and 94712 there would be one similar digit, the 7 in the 3rd digit place.) As of now I have been able to break down the inputs into the digits in each spot, I just need help comparing them. Originally I thought I could just create an int that would serve as a counter and use modulus or division to output a 1 whenever the digits were the same, but I have been unable to put together a formula that outputs a 1 or 0 depending on if the digits are alike or not.
suppose you know the length of strings n (as a condition you would need them to be equal, if they differ in length other validation is needed)
//n is the length of string
for(int i=0;i<n;i++)
{
if(string1[i]==string2[i])
{
//do something, make a counter that increments here...
//also save index i, so you can tell the position when a match occured
}else
{
//do something else if you need to do something when chars didnt match
}
}
Here you when i=0, you are comparing string1[0] with string2[0], when i=1, you compare string1[1] with string2[1] and so on.....
I'd recommend reading the two in as strings or converting to strings if you have the ability to. From there it's a simple string compare with a counter. Something like this:
#include <stdio.h>
#include <ctype.h>
#include <string.h>
int is_numeric(char *str)
{
while (*str)
if (!isdigit(*str++))
return (0);
return (1);
}
int main(void)
{
char num1[32];
char num2[32];
int count = 0;
printf("Digit 1\n>> ");
if (scanf("%5s", num1) != 1 || !is_numeric(num1))
return (0);
printf("Digit 2\n>> ");
if (scanf("%5s", num2) != 1 || !is_numeric(num2))
return (0);
if (strlen(num1) != 5 || strlen(num2) != 5)
return (0);
for (int i=0; i<5; ++i)
if (num1[i] == num2[i])
++count;
printf("%d\n", count);
return (0);
}
You can do it very easy using modulo (%) and divide (/). First you do % 10 to get the least significant digit and do the compare. Then you do / 10 to remove the least significant digit. Like:
#include <stdio.h>
#include <string.h>
int main(void) {
unsigned int i1, i2;
int i;
int cnt = 0;
printf("Input first 5 digit number:\n");
if (scanf(" %u", &i1) != 1 || i1 < 10000 || i1 > 99999) // Get integer input and check the range
{
printf("input error\n");
return 0;
}
printf("Input second 5 digit number:\n");
if (scanf(" %u", &i2) != 1 || i2 < 10000 || i2 > 99999) // Get integer input and check the range
{
printf("input error\n");
return 0;
}
for (i=0; i<5; ++i)
{
if ((i1 % 10) == (i2 % 10)) ++cnt; // Compare the digits
i1 = i1 / 10;
i2 = i2 / 10;
}
printf("Matching digits %d\n", cnt); // Print the result
return 0;
}
It can also be done using strings. Read the input as unsigned int and then convert the value to a string using snprintf and finally compare the two strings character by character.
Something like:
#include <stdio.h>
#include <string.h>
int main(void) {
char str1[32];
char str2[32];
unsigned int i1, i2;
int i;
int cnt = 0;
printf("Input first 5 digit number:\n");
if (scanf(" %u", &i1) != 1) // Get integer input
{
printf("input error\n");
return 0;
}
snprintf(str1, 32, "%u", i1);
if (strlen(str1) != 5) // Convert to string
{
printf("input error - not 5 digits\n");
return 0;
}
printf("Input second 5 digit number:\n");
if (scanf(" %u", &i2) != 1) // Get integer input
{
printf("input error\n");
return 0;
}
snprintf(str2, 32, "%u", i2); // Convert to string
if (strlen(str2) != 5)
{
printf("input error - not 5 digits\n");
return 0;
}
for (i=0; i<5; ++i)
{
if (str1[i] == str2[i]) ++cnt; // Compare the characters
}
printf("Matching digits %d\n", cnt); // Print the result
return 0;
}
The reason for taking the input into a unsigned int instead of directly to a string is that by doing that I don't have to check that the string are actually valid numbers (e.g. the user type 12W34). scanf did that for me.
The question is that show the digits which were repeated in C.
So I wrote this:
#include<stdio.h>
#include<stdbool.h>
int main(void){
bool number[10] = { false };
int digit;
long n;
printf("Enter a number: ");
scanf("%ld", &n);
printf("Repeated digit(s): ");
while (n > 0)
{
digit = n % 10;
if (number[digit] == true)
{
printf("%d ", digit);
}
number[digit] = true;
n /= 10;
}
return 0;
}
But it will show the repeated digits again and again
(ex. input: 55544 output: 455)
I revised it:
#include<stdio.h>
int main(void){
int number[10] = { 0 };
int digit;
long n;
printf("Enter a number: ");
scanf("%ld", &n);
printf("Repeated digit(s): ");
while (n > 0)
{
digit = n % 10;
if (number[digit] == 1)
{
printf("%d ", digit);
number[digit] = 2;
}
else if (number[digit] == 2)
break;
else number[digit] = 1;
n /= 10;
}
return 0;
}
It works!
However, I want to know how to do if I need to use boolean (true false), or some more efficient way?
To make your first version work, you'll need to keep track of two things:
Have you already seen this digit? (To detect duplicates)
Have you already printed it out? (To only output duplicates once)
So something like:
bool seen[10] = { false };
bool output[10] = { false };
// [...]
digit = ...;
if (seen[digit]) {
if (output[digit])) {
// duplicate, but we already printed it
} else {
// need to print it and set output to true
}
} else {
// set seen to true
}
(Once you've got that working, you can simplify the ifs. Only one is needed if you combine the two tests.)
Your second version is nearly there, but too complex. All you need to do is:
Add one to the counter for that digit every time you see it
Print the number only if the counter is exactly two.
digit = ...;
counter[digit]++;
if (counter[digit] == 2) {
// this is the second time we see this digit
// so print it out
}
n = ...;
Side benefit is that you get the count for each digit at the end.
Your second version code is not correct. You should yourself figured it out where are you wrong. You can try the below code to print the repeated elements.
#include<stdio.h>
int main(void){
int number[10] = { 0 };
int digit;
long n;
printf("Enter a number: ");
scanf("%ld", &n);
printf("Repeated digit(s): ");
while (n > 0)
{
digit = n % 10;
if (number[digit] > 0)
{
number[digit]++;;
}
else if (number[digit] ==0 )
number[digit] = 1;
n /= 10;
}
int i=0;
for(;i<10; i++){
if(number[i]>0)
printf("%d ", i);
}
return 0;
}
In case you want to print the repeated element using bool array (first version) then it will print the elements number of times elements occur-1 times and in reverse order because you are detaching the digits from the end of number , as you are seeing in your first version code output. In case you want to print only once then you have to use int array as in above code.
It is probably much easier to handle all the input as strings:
#include <stdio.h>
#include <string.h>
int main (void) {
char str[256] = { 0 }; /* string to read */
char rep[256] = { 0 }; /* string to hold repeated digits */
int ri = 0; /* repeated digit index */
char *p = str; /* pointer to use with str */
printf ("\nEnter a number: ");
scanf ("%[^\n]s", str);
while (*p) /* for every character in string */
{
if (*(p + 1) && strchr (p + 1, *p)) /* test if remaining chars match */
if (!strchr(rep, *p)) /* test if already marked as dup */
rep[ri++] = *p; /* if not add it to string */
p++; /* increment pointer to next char */
}
printf ("\n Repeated digit(s): %s\n\n", rep);
return 0;
}
Note: you can also add a further test to limit to digits only with if (*p >= '0' && *p <= '9')
output:
$./bin/dupdigits
Enter a number: 1112223334566
Repeated digit(s): 1236
Error is here
if (number[digit] == true)
should be
if (number[digit] == false)
Eclipse + CDT plugin + stepping debug - help you next time
As everyone has given the solution: You can achieve this using the counting sort see here. Time complexity of solution will be O(n) and space complexity will be O(n+k) where k is the range in number.
However you can achieve the same by taking the XOR operation of each element with other and in case you got a XOR b as zero then its means the repeated number. But, the time complexity will be: O(n^2).
#include <stdio.h>
#define SIZE 10
main()
{
int num[SIZE] = {2,1,5,4,7,1,4,2,8,0};
int i=0, j=0;
for (i=0; i< SIZE; i++ ){
for (j=i+1; j< SIZE; j++){
if((num[i]^num[j]) == 0){
printf("Repeated element: %d\n", num[i]);
break;
}
}
}
}
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.