#include<stdio.h>
#include<string.h>
void main()
{
int count,i,len;
char str[100];
printf("enter the sentence");
gets(str);
len=strlen(str);
for(i=0;i<=len;i++)
{
while(str)
if(str[i]==' ') count++;
}
printf("the number of words are :\t%d",count+1);
}
.... String line = "This is example program with spaces"; String[] tokens = line.split(" "); System.out.println(tokens.length-1); .......
public int getStringLength(String val) { return val.length(); } There is an inbuilt functionality in strings that counts the number of alphabets in a string called length()
Use the following function to count the number of digits in a string. size_t count_digits (const std::string& str) { size_t count = 0; for (std::string::const_iterator it=str.begin(); it!=str.end(); ++it) { const char& c = *it; if (c>='0' && c<='9'); ++count; } return count; }
#include<stdio.h> #include<conio.h> #include<string.h> void main() { char string[50]; int flag,count=o; clrscr(); printf("The grammar is: S->aS, S->Sb, S->ab\n"); printf("Enter the string to be checked:\n"); gets(string); if(string[0]=='a') { flag=0; for(count=1;string[count-1]!='\0';count++) { if(string[count=='b']) { flag=1; continue; } else if((flag==1)&&(string[count]=='a')) { printf("The string does not belong to the specified grammar"); break; } else if(string[count=='a']) continue; else if(flag==1)&&(string[count]='\0')) { printf("The string accepted"); break; } else { printf("String not accepted"); } getch():
#include<iostream> #include<string> size_t count_spaces(std::string& str) { size_t spaces=0; for(size_t i=0; i<str.size(); ++i) if( str[i]==32 ) ++spaces; return( spaces ); } int main() { std::string str("This is a string with some spaces."); size_t spaces = count_spaces(str); std::cout<<"The string ""<<str.c_str()<<"" has "<<spaces<<" spaces.\n"<<std::endl; }
.... String line = "This is example program with spaces"; String[] tokens = line.split(" "); System.out.println(tokens.length-1); .......
Read the characters one at a time, and write an "if" for each of the cases. In each case, if the condition is fulfilled, increment the corresponding counter variable.
public int getStringLength(String val) { return val.length(); } There is an inbuilt functionality in strings that counts the number of alphabets in a string called length()
Use the following function to count the number of digits in a string. size_t count_digits (const std::string& str) { size_t count = 0; for (std::string::const_iterator it=str.begin(); it!=str.end(); ++it) { const char& c = *it; if (c>='0' && c<='9'); ++count; } return count; }
#include<stdio.h> #include<conio.h> #include<string.h> void main() { char string[50]; int flag,count=o; clrscr(); printf("The grammar is: S->aS, S->Sb, S->ab\n"); printf("Enter the string to be checked:\n"); gets(string); if(string[0]=='a') { flag=0; for(count=1;string[count-1]!='\0';count++) { if(string[count=='b']) { flag=1; continue; } else if((flag==1)&&(string[count]=='a')) { printf("The string does not belong to the specified grammar"); break; } else if(string[count=='a']) continue; else if(flag==1)&&(string[count]='\0')) { printf("The string accepted"); break; } else { printf("String not accepted"); } getch():
#include<stdio.h> void main() { int cnt=0,i; char str[100]; printf("Enter the string "); scanf("%s",str); for(i=0;i<strlen(str)-1;i++) { if(str[i]==' ') cnt++; } printf("\nTotal no. of word in string = %d",cnt); }
Write a program to count the number of IS in any number in register B and put the count in R5.
To count the number of 'a's in a string, you can use the count() method in Python. For example, if you have a string my_string, you can get the count of 'a's by using my_string.count('a'). This will return the total number of occurrences of the letter 'a' in the string. Finally, you can print the result using the print() function.
#include<iostream> #include<string> size_t count_spaces(std::string& str) { size_t spaces=0; for(size_t i=0; i<str.size(); ++i) if( str[i]==32 ) ++spaces; return( spaces ); } int main() { std::string str("This is a string with some spaces."); size_t spaces = count_spaces(str); std::cout<<"The string ""<<str.c_str()<<"" has "<<spaces<<" spaces.\n"<<std::endl; }
#include<iostream> #include<string> using namespace std; int main() { int count=0; string b; string a[6]={"technical","school","technical","hawler","school","technical"}; for(int i=0;i<6;i++) { b=a[i]; for(int j=i;j<6;j++) { if(a[j]==b) count++; } cout<<a[i]<<" "<<count<<endl; count=0; } return 0; }
There are some ways to do it. Here I give you an example. You can do it if you take the input as string. #include <stdio.h> #include <string.h> main(void) { int i, count = 0; char ch[10000]; gets(ch); int len = strlen(ch); for(i = 0; i < len; i++) { if(ch[i] == '1') { count++; } } printf("Number of times one occurs: %d\n", count); }
You need to scan through the string and keep track of the vowelsoccurring. Here is a sample program:#include#includeint countVowels(char[] s){int count = 0, i;for( i=0; char[i] != '\0'; i++){switch(char[i]){case 'a':case 'e':case 'i':case 'u':case 'o': count++;break;}}return count;}int main(){char str[256];printf("Enter the string:\t");scanf("%s", str);printf("The number of vowels in the string are :%d\n", countVowels(str));return 0;}