I am trying to write a function to divide a string in half but after the initial input it does not output anything. My goal is to scan a year and save the first two number and the last two numbers. This is the code:
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char scan_year2() {
char year_number;
scanf("%s", year_number);
return year_number;
return 0;
}
// Function to print n equal parts of str
void divideString(char *str, int n) {
int str_size = strlen(str);
int i;
int part_size;
if (str_size % n != 0) {
printf("Invalid Input: String size");
printf(" is not divisible by n");
return;
}
part_size = str_size / 2;
for (i = 0; i < str_size; i++) {
if (i % part_size == 0)
printf("\n");
printf("%s", str[i]);
}
}
int main() {
char year_number;
scan_year2();
char str = year_number;
divideString(str, 2);
getchar();
return 0;
}
Assuming that a year is at least a 3-digit number, the best way to treat it is to treat it as a number, not as a string:
...
int year;
scanf("%d", &year);
int first = year / 100;
int last = year % 100;
printf("%d %d\n", first, last);
...
dont ignore compiler warnings, it must be complaining at you about this
char scan_year2() {
char year_number;
scanf("%s", year_number);
return year_number;
return 0;
}
you try to return twice.
Also
part_size = str_size / 2;
for (i = 0; i < str_size; i++) {
if (i % part_size == 0)
printf("\n");
printf("%s", str[i]);
}
is not going to give you the correct output. YOu are outputing the string each time. IE if str = "1923" then you will get
1923923
232
You should do
part_size = str_size / 2;
for (i = 0; i < str_size; i++) {
if (i % part_size == 0)
printf("\n");
printf("%c", str[i]);
}
to only output one char at a time
Related
Taking in input a number and then put every digit into an array, I have, converted it into a string so that I can put it into an array, but then when I use the cast to remake it a int i get the ascii number..
#include <stdio.h>
#include <stdlib.h>
int main(){
char num_str[64] = {0};
int num, cifra;
printf("Write a number: \n");
scanf("%d", &num);
int len = snprintf(num_str, 64, "%d", num);
printf("The length is %d\n", len);
for(int i = 0; i < len; i++) {
cifra = (int)(num_str[i]);
printf("%d \n", cifra);
}
return 0;
}
convert it from char to int
Test the character for digit and subtract '0'.
for(int i = 0; i < len; i++) {
int cifra = num_str[i];
if (cifra >= '0' && cifra <= '9') printf("%d\n", cifra - '0');
else printf("%c\n", cifra); // such as '-'
}
test case:
input: 1234
output: 24
input: 2468
output: 2468
input: 6
output: 6
I have this code:
#include <stdio.h>
#include <math.h>
int main() {
int num;
printf("Enter a number: \n");
scanf("%d", &num);
int numberLength = floor(log10(abs(num))) + 1;
int inputNumberArray[numberLength];
int evenNumberCount = 0;
int even[10];//new even no. array
int i = 0;
do {
inputNumberArray[i] = num % 10;
num = num / 10;
i++;
} while (num != 0);
i = 0;
while (i < numberLength) {
if (inputNumberArray[i] % 2 == 0) {
evenNumberCount ++;
even[i] = inputNumberArray[i];
}
i++;
}
printf("array count %d\n", evenNumberCount);
i = 0;
for (i = 0; i < 8; i++) {
printf(" %d", even[i]);//print even array
}
i = 0;
int result = 0;
for (i = 0; i < 10; i++) {
if (evenNumberCount == 1) {
if (even[i] != 0) {
result = even[i];
} else {
break;
}
} else {
if (even[i] != 0) {
result = result + even[i] * pow(10, i);
} else
break;
}
}
printf("\nresult is %d", result);
/*
int a = 0;
a = pow(10, 2);
printf("\na is %d", a);
*/
}
when I enter number 1234, the result/outcome is 4, instead of 24.
but when I test the rest of test case, it is fine.
the wrong code I think is this: result = result + even[i] * pow(10, i);
Can you help on this?
Thanks in advance.
why do you have to read as number?
Simplest algorithm would be
Read as text
Validate
loop through and confirm if divisible by 2 and print live
next thing, have you try to debug?
debug would let you know what are doing wrong. Finally the issue is with indexing.
evenNumberCount ++; /// this is technically in the wrong place.
even[i]=inputNumberArray[i]; /// This is incorrect.
As the user Popeye suggested, an easier approach to accomplish this would be to just read in the entire input from the user as a string. With this approach, you can iterate through each letter in the char array and use the isdigit() method to see if the character is a digit or not. You can then easily check if that number is even or not.
Here is a quick source code I wrote up to show this approach in action:
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main()
{
char input[100] = { '\0' };
char outputNum[100] = { '\0' };
// Get input from user
printf("Enter a number: ");
scanf_s("%s", input, sizeof(input));
// Find the prime numbers
int outputNumIndex = 0;
for (int i = 0; i < strlen(input); i++)
{
if (isdigit(input[i]))
{
if (input[i] % 2 == 0)
{
outputNum[outputNumIndex++] = input[i];
}
}
}
if (outputNum[0] == '\0')
{
outputNum[0] = '0';
}
// Print the result
printf("Result is %s", outputNum);
return 0;
}
I figured out the solution, which is easier to understand.
#include <stdio.h>
#include <math.h>
#define INIT_VALUE 999
int extEvenDigits1(int num);
void extEvenDigits2(int num, int *result);
int main()
{
int number, result = INIT_VALUE;
printf("Enter a number: \n");
scanf("%d", &number);
printf("extEvenDigits1(): %d\n", extEvenDigits1(number));
extEvenDigits2(number, &result);
printf("extEvenDigits2(): %d\n", result);
return 0;
}
int extEvenDigits1(int num)
{
int result = -1;
int count = 0;
while (num > 1) {
int digit = num % 10;
if (digit % 2 == 0) {
result = result == -1 ? digit : result + digit * pow(10, count);
count++;
}
num = num / 10;
}
return result;
}
}
You are overcomplicating things, I'm afraid.
You could read the number as a string and easily process every character producing another string to be printed.
If you are required to deal with a numeric type, there is a simpler solution:
#include <stdio.h>
int main(void)
{
// Keep asking for numbers until scanf fails.
for (;;)
{
printf("Enter a number:\n");
// Using a bigger type, we can store numbers with more digits.
long long int number;
// Always check the value returned by scanf.
int ret = scanf("%lld", &number);
if (ret != 1)
break;
long long int result = 0;
// Use a multiple of ten as the "position" of the resulting digit.
long long int power = 1;
// The number is "consumed" while the result is formed.
while (number)
{
// Check the last digit of what remains of the original number
if (number % 2 == 0)
{
// Put that digit in the correct position of the result
result += (number % 10) * power;
// No need to call pow()
power *= 10;
}
// Remove the last digit.
number /= 10;
}
printf("result is %lld\n\n", result);
}
}
As stated in the title I am trying to find all lower-case letters that are not in a series of words. There are no upper-case letters, digits, punctuation, or special symbols.
I need help fixing my code. I am stuck and do not know where to go from here.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(void){
int letters[26];
char words[50];
int i = 0, b = 0;
printf("Enter your input : ");
scanf("%s", words);
for(i = 0; i < 26; i++){
letters[i] = 0;
}
while(!feof(stdin)){
for(b = 0; b < strlen(words) - 1; b++){
letters[ words[b] - 'a']++;
scanf("%s", words);
}
}
printf("\nMissing letters : %c ", b + 97);
return 0;
}
My output is giving me some random letter that I do not know where it is coming from.
Here is a working first implementation.
As well as the comments that have already been made, you should use functions wherever possible to separate out the functionality of the program into logical steps. Your main function should then just call the appropriate functions in order to solve the problem. Each function should be something that is self contained and testable.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MAX_INPUT 20 /* Max input to read from user. */
char *readinput(void);
void find_missing_lower_case(char *, int);
int main()
{
char *user_input = readinput();
int len_input = strlen(user_input);
printf("user input: %s\n", user_input);
printf("len input: %d\n", len_input);
find_missing_lower_case(user_input, len_input);
/* Free the memory allocated for 'user_input'. */
free(user_input);
return 0;
}
char *readinput()
{
char a;
char *result = (char *) malloc(MAX_INPUT);
int i;
for(i = 0; i < MAX_INPUT; ++i)
{
scanf("%c", &a);
if( a == '\n')
{
break;
}
*(result + i) = a;
}
*(result + i) = '\0';
return result;
}
void find_missing_lower_case(char *input, int len_input)
{
int a = 97; /* ASCII value of 'a' */
int z = 122; /* ASCII value of 'z' */
int lower_case_chars[26] = {0}; /* Initialise all to value of 0 */
/* Scan through input and if a lower case char is found, set the
* corresponding index of lower_case_chars to 1
*/
for(int i = 0; i < len_input; i++)
{
char c = *(input + i);
if(c >= a && c <= z)
{
lower_case_chars[c - a] = 1;
}
}
/* Iterate through lower_case_chars and print any values that were not set
* to 1 in the above for loop.
*/
printf("Missing lower case characters:\n");
for(int i = 0; i < 26; i++)
{
if(!lower_case_chars[i])
{
printf("%c ", i + a);
}
}
printf("\n");
}
I figured it out and this is the code I used.
int main(void)
{
int array[26];
char w;
int i=0;
for(i=0; i<26; i++) {
array[i]=0; }
printf("Enter your input: ");
scanf("%c", &w);
while(!feof(stdin)) {
array[w-97] = 1;
scanf("%c", &w); }
printf("Missing letters: ");
for(i=0; i<26; i++) {
if(array[i] == 0) {
printf("%c ", i+97); }
}
printf("\n");
return 0;
}
I'm learning basics in coding. Can any one say what went wrong with my code
Prob:Given a string, S, of length N that is indexed from 0 to N-1 , print its even-indexed and odd-indexed characters as 2 space-separated strings on a single line
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
char input[100], final[100];
int main()
{
int num, i, j;
char even[50], odd[50], space[] = " ";
scanf("%d", &num);
for (i = 0; i < num; i++)
{
int k = 0, p = 0;
scanf(" %[^\n]s", input);
for (j = 0; input[j] != '\0'; j++)
{
if (j % 2 == 0)
{
even[k] = input[j];
k++;
}
else
{
odd[p] = input[j];
p++;
}
}
strcat(final, even);
strcat(final, space);
strcat(final, odd);
}
printf("%s", final);
}
I have to add two digit strings, meaning 1234 12+34 (at least that's what i gather). I wrote a program that does this expect for one exception, that is when the last number doesn't have a pair it wont add properly.
Here is the code i have:
void main()
{
char string[1000];
int count,sum=0,x,y;
printf("Enter the string containing both digits and alphabet\n");
scanf("%s",string);
for(count=0;count < string[count]; count++)
{
x=(string[count] - '0') * 10;
y=(string[count+1] - '0') + x;
sum += y;
count++;
}
printf("Sum of string in two digit array is =%d\n",sum);
}
so basically if i have 123 the program does 12+(30-48), instead of 12+3. Ive been sitting on it for a while, and cant figure out how to fix that issue, any tips or advice would be welcomed.
(Strings like 1234 or 4567 will do 12+34 and 45+67)
#include <stdio.h>
#include <ctype.h>
int main(void){
char string[1000];
char digits[3] = {0};
int i, j, x, sum = 0;
printf("Enter the string containing both digits and alphabet\n");
scanf("%999s", string);
for(j = i = 0; string[i]; ++i){
if(isdigit(string[i])){
digits[j++] = string[i];
if(j==2){
sscanf(digits, "%d", &x);
sum += x;
j = 0;
}
}
}
if(j==1){
digits[j] = 0;
sscanf(digits, "%d", &x);
sum += x;
}
printf("Sum of string in two digit array is = %d\n", sum);
return 0;
}