/* with run-time library */
int len;
char* str;
len = strlen (str);
/* without run-time library */
int mystrlen (const char* str) {
int len;
while (*str != '\0') {
len++;
str++;
}
return len;
}
int len;
char* str;
len = mystrlen(str);
char* cString = "Hello World!";
int length;
for (length = 0; cString[length]; ++length);
//length now holds the length of cString
else try this out!!!!!!
#include<conio.h>
#include<stdio.h>
void main()
{
char a[20];
int i;
clrscr();
gets(a);
i=0;
while(a[i]!='\0')
i++; //counts no of chars till encountering null char
printf(string length=%d,i);
getch();
}
#include<iostream.h>
#include<conio.h>
#include<stdio.h>
void main()
{
clrscr();
int c=1;
int i=0;
int b;
char d[10];
cout<<"Enter the string";
gets(d);
while(d[i]!='\0')
{
b=c++;
i++;
}
cout<<"Length"<<b;
getch();
}
***********************************************
#include<stdio.h>
#include<conio.h>
void main()
{
char a[20];
int i;
clrscr();
printf("Enter the string:");
scanf("%s",a);
for(i=0;i<20;i++)
{
if(a[i]=='\0')
{
printf("string length is:%d",i);
break;
}
}
getch();
}
***************************************************
#include<stdio.h>
#include<conio.h>
void main()
{
char a[20];
int i=0;
clrscr();
printf("Enter the string:");
scanf("%s",a);
while(a[i]!='\0')
{
i++;
}
printf("string length is:%d",i);
getch();
}
#include <windows.h> #include <stdio.h> void main() { CHAR szTest[32] = {"The Quick Brown Fox"}; size_t thelen = strlen(szTest); printf("The length of the string is %d characters\n",thelen) }
Ellipses (...) are used to show indentation for clarity.
int mystrlen(const char *p) {
... int len = 0;
... while (*p++ != '\0') len++;
... return len;
}
or (quicker):
size_t mystrlen(const char *p) {
... const char *q= p-1;
... while ((*++q)!= '\0');
... return q-p;
}
int size(const char *p) { int i; for (i=0; p[i] != '\0'; i++); return i; }
int strlen(const char* str) {
int len;
for (len=0; *str++ != '\0'; len++);
return len;
}
size_t mystrlen (const char *s)
{
register const char *t= s-1;
while (*++t);
return t-s;
}
size_t Strlen (char* pstr) {
size_t len = 0;
while (*pstr++) len++;
return len;
}
One way to do this is to convert the number to a String, then use the corresponding String method to find out the length of the String.
what is if(!(str[i]==32))
print c co com comp compu
Get the string from user , then U Split the string with space and put in a string array .Then Find Length of string array , Take first letter of each element in array, Except last. Assigned those to a string, then Fetch Last element of an array, Add it With that String.That's it.
int main (void) { char buf[1024]; scanf ("Enter a string: %s", buf); printf ("The length of the string is %d chars long.\n", strlen (buf)); return 0; }
This is an assignment not a question.
One way to do this is to convert the number to a String, then use the corresponding String method to find out the length of the String.
i am sam
what is if(!(str[i]==32))
what is if(!(str[i]==32))
No.
print c co com comp compu
You find a language that can be targeted towards the .NET Framework. What you are suggesting is something related to string manipulation and you can work with delimiters.
#include<iostream> #include<string> int main() { std::string s("The quick brown fox jumps over the lazy dog"); std::cout<<s.c_str()<<std::endl; std::cout<<"The previous string is "<<s.size()<<" characters long."<<std::endl; }
length = strlen(string);
You can't. If you want to find the length of a String object, you must use at least one of the String methods. Simply iterate over your char* and count the number of characters you find before you reach the null character . int strLength(const char* str) { int length = 0; // take advantage of the fact that all strings MUST end in a null character while( str[length] != '\0' ) { ++length; } return length; }
Get the string from user , then U Split the string with space and put in a string array .Then Find Length of string array , Take first letter of each element in array, Except last. Assigned those to a string, then Fetch Last element of an array, Add it With that String.That's it.