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3 years ago

How to write a function that counts the letters in a string in C?

Computers and Technology

1 answer:

stiv31 [10]3 years ago

3 0

Answer:

Here is the C function that counts the letters in a string:

int LetterCount(char string[]){ //function to count letters in a string passed as parameter

string[100]; // char type string with size 100

int i, letters; // letter variable stores the count for no. of letters in string

i = letters = 0; //initialize variables i and letters to 0

while (string[i] != '\0') { // loop iterates through the entire string until end of string is reached

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') ) { // if condition checks the letters in the string

letters++; } // increments 1 to the count of letters variable each time a letter is found in the string

i++; } //increments value of i to move one character forward in string

printf("Number of Letters in this String = %d", letters); // displays the number of letters in the string

return 0; }

Explanation:

Here the question means that the function should count the letters in a string. So the letters are the alphabets from a to z and A to Z.

Here is the complete program:

#include <stdio.h> // to use input output functions

int LetterCount(char string[]){ // method that takes a character string as parameter and counts the letters in the string

string[100];

int i, letters;

i = letters = 0;

while (string[i] != '\0') {

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z'))

{letters++; }

i++; }

printf("Number of Alphabets in this String = %d", letters);

return 0;}

int main(){ // start of main function

char string[100]; //declares a char array of string

printf("Please Enter a String : "); //prompts user to enter a string

fgets(string,100,stdin); //get the input string from user

LetterCount(string); } // calls method to count letters in input string

I will explain this function with an example

Lets suppose the user input "abc3" string

string[100] = "abc3"

Now the function has a while loop that while (string[i] != '\0') that checks if string character at i-th position is not '\0' which represents the end of the character string. As value of i = 0 so this means i is positioned at the first character of array i.e. 'a'

At first iteration:

i = 0

letter = 0

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') ) if condition checks if the character at i-th position of string is a letter. As the character at 0-th position of string is 'a' which is a letter so this condition evaluates to true. So the statement letter++ inside if condition executes which increments the letter variable to 1. So the value of letter becomes 1. Next statement i++ increments the value of i to 1. So i becomes 1. Hence:

i = 1

letter = 1

At second iteration:

i = 1

letter = 1

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )

if condition checks if the character at i-th position of string is a letter. As the character at 1st position of string is 'b' which is a letter so this condition evaluates to true. So the statements letter++ inside if condition and i++ executes which increments these variables to 1. Hence:

i = 2

letter = 2

At third iteration:

i = 2

letter = 2

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )

if condition checks if the character at i-th position of string is a letter. As the character at 2nd position of string is 'c' which is a letter so this condition evaluates to true. So the statements letter++ inside if condition and i++ executes which increments these variables to 1. Hence:

i = 3

letter = 3

At fourth iteration:

i = 3

letter = 3

if( (string[i] >= 'a' && string[i] <= 'z') || (string[i] >= 'A' && string[i] <= 'Z') )

if condition checks if the character at i-th position of string is a letter. As the character at 3rd position of string is '3' which is not a letter but a digit so this condition evaluates to false. So the statement letter++ inside if condition does not execute. Now i++ executes which increments this variable to 1. Hence:

i = 4

letter = 3

Now the loop breaks because while (string[i] != '\0') condition valuates to false as it reaches the end of the string.

So the statement: printf("\n Number of Letters in this String = %d", letters); executes which prints the value of letters on the output screen. Hence the output is:

Number of Letters in this String = 3

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1. Scrieţi un program care citeşte un număr natural n şi determină produsul cifrelor impare ale lui n. De exemplu, pentru n = 23

saveliy_v [14]

Answer:

num1 = input("Enter the value of n")

n = int(num1)

def calc(n):

Lst = []

while n > 0:

remainder = n % 10

Lst.append(remainder)

quotient = int(n / 10)

n = quotient

i = len(Lst) - 1

sum = 1

for i in range(0, len(Lst)):

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sum *=Lst[i];

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\ r2

num=input("Enter the value of n")

arr=[12,-12,13,15,-34,-35,35,42]

def div7positive():

sum = 0

m = 0

i = 0

while m <= 7:

if(arr[i]>=0 and arr[i]%7 == 0):

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\ r3.Write an algorithm that reads a natural number n and calculates the sum:

\ rS = 1 / (1 * 2) + 1 / (2 * 3) + 1 / (3 * 4) +… + 1 / ((n-1) * n)

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num=input("Enter the value of n")

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print("sum of series is", +sum)

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\ R4. Read a natural number n. Calculate the sum of its own divisors n. For example, for n = 12, the sum of its own divisors is 2 + 3 + 4 + 6 = 15

num=input("Enter the value of n")

def sumdivisors():

sum = 0

m= 2

while m <= int(num):

if int(num) % m == 0:

sum += m

m += 1

print("sum of divisors is", +sum)

sumdivisors()\ r

\ R5. We read a natural number n and then whole numbers. Calculate and display the sum of the natural numbers between 10 and 100. For example, if n = 5 and then read 30, –2, 14, 200, 122, then the sum will be 44 (that is, 30 + 14).

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Lst = []

num=input("Enter the value of n")

i = 0

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Lst.append(int(num1));

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print("sum of numbers between 0 and 100 is", +sum)

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\ r

num = input("Enter the value of n")

n = int(num)

print(“sum as required is:” (n -9) + n)

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num = input("Enter the value of n")

n = int(num)

Lst = []

def calc(n):

i = 1

total = pow(10,n)

while i <= total:

j = i

while j > 0:

remainder = j % 10

Lst.append(remainder)

quotient = int(j / 10)

if quotient > 0:

j = quotient

else:

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while length >= 0:

sum += Lst[length]

length = length - 1

if sum == pow(10,n):

print(j)

k = len(Lst)

del Lst[0:k]

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return(0)

calc(n)

\ r

\ R8. We consider the row 1, 1, 2, 3, 5, 8, 13, ... in which the first two terms are 1, and any other term is obtained from the sum of the preceding two. Write an algorithm that reads a natural number n and displays the first n terms of this string. For example, for n = 6, 1, 1, 2, 3, 5, 8 will be displayed.

\ r

nterm = int(input("What number of terms do you want?"))

a, b = 0, 1

totalcount = 0

if nterm <= 0:

print("Please input a positive number")

elif nterm == 1:

print("Fibonacci number upto which",nterm,":")

print(a)

else:

print("Fibonacci series:")

print(nterm)

while totalcount < nterm:

print(a)

nth = a + b

a = b

b = nth

totalcount += 1

\ 9. Write an algorithm that reads two natural numbers n1 and n2 and displays the message "yes" if the sum of the squares of the digits of n1 is equal to the sum of the numbers of n2 or "no" otherwise. For example, for n1 = 232 and n2 = 881, "yes" will be displayed, and for n1 = 45 and n2 = 12, "no" will be displayed.

num1 = input("Enter the value of n")

num3 = input("Enter the value of n")

n = int(num1)

num2 = int(num3)

def calc(n):

Lst = []

while n > 0:

remainder = n % 10

Lst.append(remainder)

quotient = int(n / 10)

n = quotient

i = len(Lst) - 1

sum = 0

while i >= 0:

sum += Lst[i]* Lst[i]

i = i - 1

return(sum)

def calc1(num2):

Lst = []

while num2 > 0:

remainder = num2 % 10

Lst.append(remainder)

quotient = int(num2 / 10)

num2 = quotient

i = len(Lst) - 1

sum = 0

while i >= 0:

sum = Lst[i] + Lst[i]

i = i - 1

return(sum)

a = calc(n)

b = calc1(num2)

if a == b:

print("yes")

else:

print("No")

\ r

\ R10. Write an algorithm that reads a natural number n and displays the message "yes" if all of its n numbers are distinct, or "no" if n does not have all the distinct digits. For example, for n = 37645 it will display "yes" and for 23414 it will show "no".

\ r

num1 = input("Enter the value of n")

n = int(num1)

def calc(n):

Lst = []

while n > 0:

remainder = n % 10

Lst.append(remainder)

print( remainder)

quotient = int(n / 10)

n = quotient

flag = 1

for i in range(0, len(Lst)):

for j in range(i+1, len(Lst)):

if Lst[i] == Lst[j]:

flag = 0

break

if flag == 1:

print("YES")

else:

print("NO")

return(0)

Explanation:

Please check answer.

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3 years ago

Which of these purchases is most likely to be paid for with a credit card

Vesna [10]

Answer:

plane ticket?

Explanation:

7 0

3 years ago

toThePowerOf is a function that accepts two int parameters and returns the value of the first parameter raised to the power of t

zlopas [31]

Answer:

cubeVolume = toThePowerOf(cubeSide, 3);

Explanation:

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