Using spaces while taking integers from the user [duplicate] - arrays

Question

This question already has answers here:
Reading string from input with space character? [duplicate]
(13 answers)
Closed last year.
I'm trying to practice some stuff about text files, printing and reading from them. I need to take input from user -maybe their phone number or someone else's- but i want them to be able to use spaces between numbers
Lets say my phone number is: 565 856 12
i want them to be able to give me this number with spaces, instead of a squished version like 56585612
So far i've tried scanf() and i don't know how to make scanf() do something like this. I've tried going for chars and for loops but its a tangle.
And when i type 565 856 12 and press enter, only 565 will be counted for the phone number. and 856 12 goes for the next scanf.
struct Student{
unsigned long long student_phone_number;
}
int main(){
FILE *filePtr;
filePtr = fopen("std_info.txt","w");
struct Student Student1;
printf("\nEnter Student's Phone Number: ");
scanf("%llu",&Student1.student_phone_number);
fprintf(filePtr,"%llu\t",Student1.student_phone_number);
}

Answer 1

To solve this problem, I modified the Student structure to store both unsigned long long integers and character array. User reads character array from stdin. The read data is validated using the isValid() method, and the string is converted to an unsigned long long integer using the convertToNumber() method.
#include <stdio.h>
#include <ctype.h>
#include <stdbool.h>
struct Student{
unsigned long long numberForm;
char *textForm;
};
// Converts character array to unsigned long long integer.
void convertToNumber(struct Student * const student);
// Validates the data in the Student.textForm variable.
bool isValid(const char * const input, const char * const format);
// Returns the number of characters in a character array.
size_t getSize(const char * const input);
// This function returns the "base ^ exponent" result.
unsigned long long power(int base, unsigned int exponent);
int main()
{
struct Student student;
char format[] = "NNN NNN NN"; /* "123 456 78" */
printf("Enter phone number: ");
fgets(student.textForm, getSize(format) + 1, stdin);
// The gets() function is deprecated in newer versions of the C/C++ standards.
if(isValid(student.textForm, format))
{
convertToNumber(&student);
printf("Result: %llu", student.numberForm);
}
return 0;
}
void convertToNumber(struct Student * const student)
{
int size = getSize(student->textForm) - 2;
unsigned int temp[size];
student->numberForm = 0ull;
for(int i = 0, j = 0 ; i < getSize(student->textForm) ; ++i)
if(isdigit(student->textForm[i]))
temp[j++] = student->textForm[i] - '0';
for(size_t i = 0 ; i < size ; ++i)
student->numberForm += temp[i] * power(10, size - i - 1);
}
bool isValid(const char * const input, const char * const format)
{
if(getSize(input) == getSize(format))
{
size_t i;
for(i = 0 ; i < getSize(input) ; ++i)
{
if(format[i] == 'N')
{
if(!isdigit(input[i]))
break;
}
else if(format[i] == ' ')
{
if(input[i] != format[i])
break;
}
}
if(i == getSize(input))
return true;
}
return false;
}
unsigned long long power(int base, unsigned int exponent)
{
unsigned long long result = 1;
for(size_t i = 0 ; i < exponent ; ++i)
result *= base;
return result;
}
size_t getSize(const char * const input)
{
size_t size = 0;
while(input[++size] != '\0');
return size;
}
This program works as follows:
Enter phone number: 123 465 78
Result: 12346578

Answer 2

You can use fgets to parse an input with spaces and all:
#include <stdio.h>
#define SIZE 100
int main() {
char str[SIZE];
fgets(str, sizeof str, stdin); // parses input string with spaces
// and checks destination buffer bounds
}
If you then want to remove the spaces you can do that easily:
#include <ctype.h>
void remove_white_spaces(char *str)
{
int i = 0, j = 0;
while (str[i])
{
if (!isspace(str[i]))
str[j++] = str[i];
i++;
}
str[j] = '\0';
}
Presto, this function will remove white spaces from the string you pass as an argument.
Live demo
Input:
12 4 345 789
Output:
124345789
After that it's easy to convert this into an unsigned integral value, you can use strtoul, but why would you store a phone number in a numeric type, would you be performing arithmetic operations on it? Doubtfully. And you then save to a file, so it really doesn't matter if it's a string or numeric type. I would just keep it as a string.

Answer 3

It is generally better to store a telephone number as a string, not as a number.
In order to read a whole line of input (not just a single word) as a string, I recommend that you use the function fgets.
Here is an example based on the code in your question:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct Student{
char phone_number[50];
};
int main( void )
{
//open output file (copied from code in OP's question)
FILE *filePtr;
filePtr = fopen("std_info.txt","w");
//other variable declarations
struct Student Student1;
char *p;
//prompt user for input
printf( "Enter student's phone number: ");
//attempt to read one line of input
if ( fgets( Student1.phone_number, sizeof Student1.phone_number, stdin ) == NULL )
{
printf( "input error!\n" );
exit( EXIT_FAILURE );
}
//attempt to find newline character, in order to verify that
//the entire line was read in
p = strchr( Student1.phone_number, '\n' );
if ( p == NULL )
{
printf( "line too long for input!\n" );
exit( EXIT_FAILURE );
}
//remove newline character by overwriting it with terminating
//null character
*p = '\0';
//write phone number to file
fprintf( filePtr, "%s\n", Student1.phone_number );
//cleanup
fclose( filePtr );
}

Answer 4

Use string for that purpose and trim the spaces from string and then convert the given number into integer using atoi function, to use atoi you must include stdlib.h header file. For example
#include<stdio.h>
#include<stdlib.h>
int main(){
char str[] = "123456";
unsigned long long int num = atoi(str);
printf("%llu", num);
}