Do you know how I can detect space and letters in a CHAR variable?
I need to detect letters or space in a input of numbers:
This what I want to do:
Enter Document Number of 8 numbers:
// i press space and pressed enter
ERROR: please enter the age again: 4fpdpfsg
There's where my code doesn't detect the letters after the 4, and what I want is recognize that there's letters in the input, and then shows only the 4.
int isLetter(char input[]){
int i = 0;
while(input[i]!='\0'){
if((input[i]!=' ') && (input[i]<'a'||input[i]>'z') && (input[i]<'A'||input[i]>'Z'))
return 0;
i++;
}
return 1;
}
The standard C library has various character type testing functions. They are declared in the #include <ctype.h> header.
Unfortunately, the obvious way of using these functions is often wrong. They take an argument of type int which is actually expected to be an unsigned character value (a byte, effectively) in the range 0 to UCHAR_MAX. If you pass in a char value which happens to be negative, undefined behavior ensues, which might work by coincidence, crash or worse yet form a vulnerability similar to heartbleed (possibly worse).
Therefore the cast to (unsigned char) is quite likely necessary in the following:
#include <ctype.h>
/* ... */
char ch;
/* ... */
if (isalpha((unsigned char) ch) || ch == ' ') {
/* ch is an alphabetic character, or a space */
}
Simple character constants (not numeric escaped ones) derived from the C translation time character set have positive values in the execution environment; code which can safely assume that it only manipulates such characters can do without the cast. (For instance, if all the data being manipulated by the program came from string or character literals in the program itself, and all those literals use nothing but the basic C translation time character set.)
That is to say, isalpha('a') is safe; a is in the C translation time character set, and so the value of the character constant 'a' is positive. But say you're working with source code in ISO-8859-1 and have char ch = 'à';. If char is signed, this ch will have a negative value, which is fine according to ISO C because an accented à isn't in the basic C translation character set. The expression isalpha(ch); then passes a negative value to the isalpha function, which is wrong.
Try:
if (!((input[i] == ' ') || (input[i] >= 'a' && input[i] <= 'z') || (input[i] >= 'A' && input[i] <= 'Z')))
or, better:
#include <ctype.h>
if (!((input[i] == ' ') || isalpha(input[i])))
You could use sscanf(input,"%d%n",&number,&nrOfDigits) which reads in an integral value into number and additionally stores the position of the first character which has not been part of the number in nrOfDigits. With this information, you can then decide what to do, e.g. nrOfDigits < 8 would indicate that either the input was shorter than 8 characters, or that it does contain less than 4 consecutive digits. See sample code of the usage below.
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int isLetter(char input[]){
int nrOfDigits=0;
int number;
int scannedElems = sscanf(input,"%d%n",&number,&nrOfDigits);
if (scannedElems == 0) {// number could not be read--
printf ("No number read.\n");
return 0;
}
else {
char c = input[nrOfDigits];
int isAlpha = isalpha(c);
printf("input %s leads to number %d with %d digit(s); first characer after the digits is '%c', (isalpha=%d)\n", input, number, nrOfDigits, c, isAlpha);
return number;
}
}
int main(){
isLetter("4fpdpfsg"); // input 4fpdpfsg leads to number 4 with 1 digit(s); first characer after the digits is 'f', (isalpha=1)
isLetter("afpdpfsg"); // No number read.
isLetter("12345678"); // input 12345678 leads to number 12345678 with 8 digit(s); first characer after the digits is '�', (isalpha=0)
return 0;
}
BTW: you could implement a similar logic with strtoul as well.
hey guys i finally get the way to detect the input is conformed only for 8 numbers theres the code
char* InputDni(char dni[])
{
int sizeletter;
int i;
fflush(stdin);
gets(dni);
// 8 is the size of DNI in argentina
while((isLetter(dni)) || (strlen(dni)!=8))
{
printf("ERROR: enter again the DNI: ");
fflush(stdin);
gets(dni);
}
sizeletter=strlen(dni);
for(i=0 ;i<sizeletter; i++)
{
while(isalpha(dni[i]))
{
printf("ERROR: enter again the DNI: ");
fflush(stdin);
gets(dni);
i++
}
}
return dni;
}
//isLetter
int isLetter(char input[])
{
int i = 0;
int sizeletter;
int flag=1;
sizeletter=strlen(input);
for(i=0;i<sizeletter;i++)
{
if((input[i]!=' ') && (input[i]<'a'||input[i]>'z') && (input[i]<'A'||input[i]>'Z'))
{
flag=0;
}
}
return flag;
}
picture of the code running in cmd:
Related
Given a string containing alphanumeric characters, calculate the sum of all numbers present in the string.
The problem with my code is that it displays the integers present before the characters, but it is not summing up the integers after the characters.
The execution is easy in python and C++ but I cant get it done using C! Can anyone please verify where I have done wrong? << thank you !
enter code here
#include<stdio.h>
#include<string.h>
int convert(char[]);
int main()
{
char ch[100],temp[100]={0};
int i=0,s=0,j=0,n;
scanf("%s",ch);
for(i=0;i<strlen(ch);i++)
{
if((ch[i]>='0') && (ch[i]<='9'))
{
temp[j]=ch[i];
j++;
}
else
{
if(temp[0]== '\0')
{
continue;
}
else
{
n=convert(temp);
s+=n;
temp[0]= '\0';
j=0;
}
}
}
printf("%d",s);
return 0;
}
int convert(char s[]) //converting string to integer
{
int n=0;
for(int i=0;i<strlen(s);i++)
{
n= n * 10 + s[i] - '0';
}
return n;
}
Input : 12abcd4
Expected output : 16
But the output is 12 for my code.
There are two problems in your code. The first was mentioned in the comments : if the last character is a digit, the last "number section" will not be taken into account. But I don't think that the solution given in the comments is good because if the last character is not a digit, you will have a wrong value. To correct this, I added an if statement that check if the last character is a digit, if so call convert().
The second problem is that strlen return the number of characters in you string from the beginning until it finds an '\0'. The way you used your string lead to the follow problem :
ch = "12abcd4".
At first you have temp = '1' + '2' + '\0'...
After calling convert() you set temp[0] to '\0', thus temp = '\0' + '2' + '\0'... .
And when you start reading digit again, you set '4' in temp[0]. Your string is now : '4' + '2' + '\0'... .
The n returned will be 42 and your result 54 (12+42). There are several solution to have the expected behavior, I chose to use your variable j to indicate how many characters should be read instead of using strlen() :
#include<stdio.h>
#include<string.h>
int convert(char[], int size);
int main() {
char ch[100],temp[100]={0};
int i=0,s=0,j=0,n;
scanf("%s",ch);
for(i=0;i<strlen(ch);i++) {
if((ch[i]>='0') && (ch[i]<='9')) {
temp[j]=ch[i];
j++;
// change here
if(i == strlen(ch) - 1) {
n=convert(temp, j);
s+=n;
}
}
else {
// change here
n=convert(temp, j);
s+=n;
if(temp[0]== '\0') {
continue;
}
temp[0]= '\0';
j=0;
}
}
printf("%d\n",s);
return 0;
}
//change here
int convert(char s[], int size) {
int n=0;
for(int i=0;i<size;i++) {
n= n * 10 + s[i] - '0';
}
return n;
}
You could use a combination of strtoul() and strpbrk() to do this.
Declare two character pointers start_ptr and end_ptr and make start_ptr point to the beginning of the string under consideration.
char *start_ptr=s, *end_ptr;
where s is the character array of size 100 holding the string.
Since your string has only alphanumeric characters, there is no - sign and hence there are no negative numbers. So we can get away with using unsigned integers.
We are using strtoul() from stdlib.h to perform the string to integer conversion. So let's declare two variables: rv for holding the value returned by strtoul() and sum to hold the sum of numbers.
unsigned long rv, sum_val=0;
Now use a loop:
for(; start_ptr!=NULL; )
{
rv = strtoul(start_ptr, &end_ptr, 10);
if(rv==ULONG_MAX && errno==ERANGE)
{
//out of range!
printf("\nOut of range.");
break;
}
else
{
printf("\n%lu", rv);
sum_val += rv;
start_ptr=strpbrk(end_ptr, "0123456789");
}
}
strtoul() will convert as much part of the string as possible and then make end_ptr point to the first character of the part of the string that could not be converted.
It will return ULONG_MAX if the number is too big and errno would be set to ERANGE.
Otherwise the converted number is returned.
strpbrk() would search for a set of characters (in this case the characters 0-9) and return a pointer to the first match. Otherwise NULL is returned.
Don't forget to include the following header files:
stdlib.h ---> strtoul
string.h ---> strpbrk
limits.h ---> ULONG_MAX
errno.h ---> errno
In short, we could make the program to something like
for(; start_ptr!=NULL; sum_val += rv, start_ptr=strpbrk(end_ptr, "0123456789"))
{
rv = strtoul(start_ptr, &end_ptr, 10);
if(rv==ULONG_MAX && errno==ERANGE)
{
//out of range!
break;
}
}
printf("\n\n%lu", sum_val);
So the value of sum_val for the string "12abcd4" would be 16.
scanf() is usually not the best way to accept input that is not well-formatted. Maybe you can use fgets()-sscanf() combo instead.
If you must use scanf(), make sure that you check the value returned by it, which in your case must be 1 (the number of successful assignments that scanf() made).
And to prevent overflow, use a width specifier as in
scanf("%99s",ch);
instead of
scanf("%s",ch);
as 100 is the size of the ch character array and we need one extra byte to store the string delimiter (the \0 character).
#include "stdafx.h"
#include "stdlib.h"
#include <ctype.h>
int num = 0;
int i = 0;
int ch = 0;
int letter_index_in_alphabet(int ch) {
if (isalpha(ch) == true) {
char temp_str[2] = { ch };
num = strtol(temp_str, NULL, 36) - 9;
printf("%d is a letter, with %d as its location in the alphabet!", ch, num);
}
else {
return -1;
}
}
int main()
{
char input_str[10];
printf("Please enter a series of up to 10 letters and numbers: \n");
fgets(input_str, 10, stdin);
for (i == 0; i <= 10; i++) {
ch = input_str[i];
letter_index_in_alphabet(ch);
}
return 0;
}
Hello everyone, this is my first post on SOF! The goal of this program is to read characters from the standard input to EOF. For each character, report if it is a letter. If it is a letter, print out its respective index in the alphabet ('a' or 'A' = 1, 'b' or 'B' = 2..etc). I have been searching some other posts on stackoverflow and this has helped me get this far(using fgets and strtol functions). I have no visible syntax errors when I run this code, but after I enter a string of characters (ex: 567gh3fr) the program crashes.
Basically, I am trying to use 'fgets' to bring each character entered into a string with the appropriate index. Once I have that string, I check each index for a letter and if it is, I print the number assigned to that letter of the alphabet.
Any help or insight into why this isn't working as intended is greatly appreciated, Thanks!
You have a few problems.
First, char input_str[10] is only big enough for the user to enter 9 characters, not 10, because you need to allow one character for the null byte that ends a string.
Second, your loop goes too far. For a string with 10 characters, indexes go up to 9, not 10. It also should stop when it gets to the null byte, since the user might not have entered all 9 characters.
To get the position in the alphabet, you can simply subtract the value of A or a from the value of the character. Use tolower() or toupper() to convert the character to the case that you're going to use. Your method works, but it's overly complicated and confusing.
letter_index_in_alphabet() is declared to return int. But when the character is a letter, it doesn't execute a return statement. I'm not sure why it's supposed to return something, since you never use the return value, but I've changed it to return the position (maybe the caller should be the one that prints the message, so the function just does the calculation).
In the for loop, it should be i = 0 to perform an assignment, not i == 0 which is comparison.
You also shouldn't use global variables so much. And system header files should have <> around them, not "".
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <ctype.h>
int letter_index_in_alphabet(int ch) {
if (isalpha(ch)) {
int num = tolower(ch) - 'a' + 1;
printf("%d is a letter, with %d as its location in the alphabet!\n", ch, num);
return num;
} else {
return -1;
}
}
int main()
{
char input_str[10];
printf("Please enter a series of up to 9 letters and numbers: \n");
fgets(input_str, sizeof(input_str), stdin);
for (int i = 0; input_str[i]; i++) {
letter_index_in_alphabet(input_str[i]);
}
return 0;
}
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I have to write a program that converts an user input (which is a string) to an Integer. In the same time it should check, if the user input really is a number.
And also everything in one method.
and NO LIBRARY FUNCTIONS allowed.
I can't figure any idea how to do it. All I got for the beginning is just this pathetic structure
#include <stdio.h>
void main()
{
char input[100];
int i;
int sum = 0;
printf("Type a String which will be converted to an Integer: ");
scanf("%c, &input");
for (i = 0; i < 100; i++)
{
}
}
I appreciate any help, thanks
The conversion is the easy part...
But if you must not use library functions,
there is only one way to take a string, and that is argv;
there is only one way to give an integer, and that is the exit code of the program.
So, without much ado:
int main( int argc, char * argv[] )
{
int rc = 0;
if ( argc == 2 ) // one, and only one parameter given
{
unsigned i = 0;
// C guarantees that '0'-'9' have consecutive values
while ( argv[1][i] >= '0' && argv[1][i] <= '9' )
{
rc *= 10;
rc += argv[1][i] - '0';
++i;
}
}
return rc;
}
I did not implement checking for '+' or '-', and did not come up with a way to signal "input is not a number". I also just stop parsing at the first non-digit. All this could probably be improved upon, but this should give you an idea of how to work around the "no library functions" restriction.
(Since this sounds like a homework, you should have to write some code of your own. I already gave you three big spoons of helping regarding argv, the '0'-'9', and the conversion itself.)
Call as:
<program name> <value>
(E.g. ./myprogram 28)
Check return code with (for Linux shell):
echo $?
On Windows it's something about echo %ERRORLEVEL% or somesuch... perhaps a helpful Windows user will drop a comment about this.
Source for the "'0'-'9' consecutive" claim: ISO/IEC 9899:1999 5.2.1 Character sets, paragraph 3:
In both the source and execution basic character sets, the value of each character after 0 in the above list of decimal digits shall be one greater than the value of the previous.
I'm sure this is preserved in C11, but I only have the older C99 paper available.
Take hightes digit and add it to number, multiply the number by 10 and add the next digit. And so on:
#include <stdio.h> // scanf, printf
void main()
{
char input[100];
printf("Type a String which will be converted to an Integer: ");
scanf("%s", input);
int number = 0;
int neg = input[0] == '-';
int i = neg ? 1 : 0;
while ( input[i] >= '0' && input[i] <= '9' )
{
number *= 10; // multiply number by 10
number += input[i] - '0'; // convet ASCII '0'..'9' to digit 0..9 and add it to number
i ++; // step one digit forward
}
if ( neg )
number *= -1;
printf( "string %s -> number %d", input, number );
}
input[i] - '0' works, because ASCII characters '0'..'9' have ascending ASCII codes from 48 to 57.
So basically you want to know how something like the standard library atoi works. In order to do this, you need to consider how strings represent numbers.
Basically, a string (that represents a number) is a list o digits from 0 to 9. The string abcd (where a, b, c, d are placeholders for any digit) represents the number a*10 ^ 3 + b*10^2 + c * 10 + d (considering base 10 here, similar for other bases). So basically you need to decompose the string as shown above and perform the required arhitmetic operations:
// s - the string to convert
int result = 0;
for (int index = 0; index < strlen(s); index++) {
result = result * 10 + s[index] - '0';
}
The operation s[index] - '0' converts the character that represent a digit to its value.
// the function returns true for success , and false for failure
// the result is stored in result parameter
// nb: overflow not handled
int charToInt(char *buff,int *result){
*result=0;
char c;
while(c=*buff++){
if((c < '0') || (c >'9')) // accept only digits;
return 0;
*result *= 10;
*result += c-'0';
}
return 1;
}
Lot of things which are missed. Firstly taking a string in is done by scanf("%s",input); By the way in which you are receiving it, it only stores a character, secondly run the loop till the length of the string recieved. Check the below code.
#include <stdio.h>
#include<string.h>
void main()
{
char input[100];
int i;
int sum = 0;
printf("Type a String which will be converted to an Integer: ");
scanf("%s", input);
for (i = 0; i < strlen(input); i++)
{
if(input[i]>=48 && input[i]<=57)
{
//do something, it is a digit
printf("%d",input[i]-48);
//48 is ascii value of 0
}
}
Try it:
#include <stdio.h>
void main()
{
char input[100];
int i,j;
int val = 0;
printf("Type a String which will be converted to an Integer: ");
scanf("%s",input);
for(j=0; input[j] != '\0'; j++); // find string size
for (i = 0; i < j; i++)
{
val = val * 10 + input[i] - 48;
}
}
If you want your code to be portable to systems that don't use ASCII, you'll have to loop over your char array and compare each individual character in the source against each possible number character, like so:
int digit;
switch(arr[i]) {
case '0':
digit=0; break;
case '1':
digit=1; break;
// etc
default:
// error handling
}
Then, add the digit to your result variable (after multiplying it by 10).
If you can assume ASCII, you can replace the whole switch statement by this:
if(isdigit(arr[i])) {
digit=arr[i] - '0';
} else {
// error handling
}
This works because in the ASCII table, all digits are found in a single range, in ascending order. By subtracting the ordinal value of the zero character, you get the value of that digit. By adding the isdigit() macro, you additionally ensure that only digit characters are converted in this manner.
I am a newbie to C and I was looking over some questions where I pondered upon a question where we need to scan in values using the users input. Example
1 2 3 45 6 7. So Automatically we scan these values into a 2D array.
One thing that troubles me is what If the user inputs
1 2 3 2 3 Josh, how can we ignore Josh and only scan in the values into the array.
I looked at using getchar and use a flag variable but I am unable to figure out the conundrum of differentiating between the integer and character.
/* This is something that I tried */
#include <stdio.h>
int main(int argc, char *argv[]) {
int a;
int b;
int A[10];
while (((a = getchar()) != '\n') && (b = 0)) {
if (!(a >= "A" && a <= "Z")) {
scanf("%d", A[b]);
}
b++;
}
}
}
I think one good method for achieving what you want is using scanf with the format "%s", which will read everything as a string, effectively splitting the input according to white spaces. From the manual:
s
Matches a sequence of non-white-space characters; the next
pointer must be a pointer to character array that is long
enough to hold the input sequence and the terminating null
byte ('\0'), which is added automatically. The input string
stops at white space or at the maximum field width, whichever
occurs first.
To convert the string to integer, you can use atoi. From the manual:
The atoi() function converts the initial portion of the string
pointed to by nptr to int.
So, if it converts the initial portion of the string into an integer, we can use that to identify what is a number and what's not.
You can build a simple "word detector" for atoi.
Using the function isalpha from ctype.h you can do:
int isword(char *buffer)
{
return isalpha(*buffer);
}
And rewriting your reading program you have:
#include <stdio.h>
#include <ctype.h>
int isword(char *buffer)
{
return isalpha(*buffer);
}
int main(void)
{
char input[200];
int num;
while (1) {
scanf("%s", input);
if (!strcmp(input, "exit")) break;
if (isword(input)) continue;
num = atoi(input);
printf("Got number: %d\n", num);
}
return 0;
}
You should keep in mind that the name isword is fallacious. This function will not detect if buffer is, in fact, a word. It only tests the first character and if that is a character it returns true. The reason for this is the way our base function itoa works. It will return zero if the first character of the buffer is not a number - and that's not what you want. So, if you have other needs, you can use this function as a base.
That's also the reason I wrote a separate function and not:
if (!isalpha(input[0]))
num = itoa(input);
else
continue;
The output (with your input):
$ ./draft
1 2 3 2 3 Josh
Got number: 1
Got number: 2
Got number: 3
Got number: 2
Got number: 3
exit
$
About assigments and &&
while (((a = getchar()) != '\n') && (b = 0))
As I said in a comment, this loop will never work because you're making a logical conjunction(AND) with an assignment that will always return zero. That means the loop condition will always evaluate to false.
In C, assignments return the value assigned. So, if you do
int a = (b = 10);
a will have now hold the value 10. In the same way, when you do
something && (b = 0)
You're effectively doing
something && 0
Which will always evaluate to false (if you remember the AND truth table):
p q p && q
---------------
0 0 0
0 1 0
1 0 0
1 1 1
Your code is completely wrong. I suggest to delete it.
You could use scanf with %d to read in numbers. If it returns 0, there is some invalid input. So, scan and discard a %s and repeat this process:
int num = -1;
while(num != 0)
{
printf("Enter a number, enter 0 to exit:");
if(scanf("%d", &num) == 0) /* If scanf failed */
{
printf("Invalid input found!");
scanf("%*s"); /* Get rid of the invalid input (a word) */
}
}
I'm coding a basic program to check if a string is a palindrome or not.
#include <stdio.h>
#include <string.h> //Has some very useful functions for strings.
#include <ctype.h> //Can sort between alphanumeric, punctuation, etc.
int main(void)
{
char a[100];
char b[100]; //Two strings, each with 100 characters.
int firstchar;
int midchar;
int lastchar;
int length = 0;
int counter = 0;
printf(" Enter a phrase or word for palindrome checking: \n \n ");
while ((a[length] == getchar()) !10 ) //Scanning for input ends if the user presses enter.
{
if ((a[length -1]), isalpha) // If a character isalpha, keep it.
{
b[counter] = a[length-1];
counter++;
}
length--; //Decrement.
}
makelower(b, counter); //Calls the function that changes uppercase to lowercase.
for( firstchar = 0; firstchar < midchar; firstchar++ ) //Compares the first and last characters.
{
if ( a[firstchar] != a[lastchar] )
{
printf(", is not a palindrome. \n \n");
break;
}
lastchar--;
}
if( firstchar == midchar )
{
printf(", is a palindrome. \n \n");
}
return 0;
}
//Declaring additional function "makelower" to change everything remaining to lowercase chars.
int makelower (char c[100], int minicount)
{
int count = 0;
while (count <= minicount)
{
c[count] = tolower(c[count]);
}
return 0;
}
And I'm getting the following compiler error on the line with the first while loop, immediately after the printf statement:
p5.c: In function 'main':
p5.c:30: error: expected ')' before '!' token
I've looked up and down, but I haven't found any out-of-place or nonpartnered parenthesis. The only thing I can think of is that I'm missing a comma or some kind of punctuation, but I've tried placing a comma in a few places to no avail.
Sorry if this is too specific. Thanks in advance.
while ((a[length] == getchar()) !10 )
What it looks like you're trying for is assigning to a[length] the result of getchar() and verifying that that is not equal to 10. Which is spelled like so:
while ((a[length] = getchar()) != 10)
= is assignment, == is the test.
Further, your counters are confused. length is initialized to 0 and is only decremented, which will lead to falling off the front of the array after the first decrement. This doesn't get a chance to happen, because you attempt to access a[length-1], which will also fail. This looks like a off-by-one error, also known as a fencepost error, in accessing the character you just read from getchar().
Also, since nothing is checking that the length of recorded input doesn't exceed the length of your buffer a[100], you could fall off the end there as well.
The counters for your palindrome check function are also off. midchar and lastchar are never initialized, midchar is never set, and lastchar is decremented without ever having a value set. You would probably be better off testing a[firstchar] == a[(counter-1)-firstchar].