 How to compare literals in COUNTIF
By : user1726724
Date : March 29 2020, 07:55 AM
this will help Named lists in an excel sheet are referenced from another sheet on the same book. code :
COUNTIF(Range, "=> 5")
|
 C++ compare two string literals
By : Akash Tyagi
Date : March 29 2020, 07:55 AM
To fix this issue When comparing a string literal with another string literal with the == operator (or !=), is the result well defined? , code :
"a" == "a"
"a" != "b"
|
 Integer literals
By : Ujjwal Bhaumik
Date : March 29 2020, 07:55 AM
I wish this help you You cannot assign a value of a type with higher capacity to a type of lower capacity without casting. long can store larger numbers, so it can't be assigned to int.
|
 I am trying to compare string literals and I want to remove repeated literals I want to do it without using POINTERS.,
By : Shruti Chandak
Date : March 29 2020, 07:55 AM
wish helps you I am trying to compare string literals and I want to remove repeated literals I want to do it without using POINTERS. , Correcting your code: code :
#include <stdio.h>
#include <string.h>
int main()
{
char str[30];
printf("Enter strings : ");
fgets(str,30,stdin);
char tem[30];
size_t count;
size_t county=0;;
for(count = 0 ; count < strlen(str)-1 ; count++) {
if(str[count] != str[count+1]) {
tem[county++] = str[count];
}
}
tem[county] = '\0';
printf("%s\n", tem);
return 0;
}
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
int main()
{
char str[30];
printf("Enter strings : ");
fgets(str,30,stdin);
char tem[30];
size_t count;
size_t county;
size_t tem_index = 0;
size_t size_of_string = strlen(str);
bool found;
for(count = 0 ; count < size_of_string-1; count++)
{
found = false;
county = count+1;
while ((found == false) && (county<size_of_string))
{
if(str[count] == str[county])
{
found = true;
}
county++;
}
if (found == false)
{
tem[tem_index++] = str[count];
}
}
tem[tem_index] = '\0';
printf("%s\n", tem);
return 0;
}
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
int main()
{
char str[30];
printf("Enter strings : ");
fgets(str,30,stdin);
char tem[30];
size_t count;
size_t county;
size_t tem_index = 0;
size_t size_of_string = strlen(str);
bool found;
for(count = 0 ; count < size_of_string-1; count++)
{
found = false;
county = 0;
while ((found == false) && (county<tem_index))
{
if(str[count] == tem[county])
{
found = true;
}
county++;
}
if (found == false)
{
tem[tem_index++] = str[count];
}
}
tem[tem_index] = '\0';
printf("%s\n", tem);
return 0;
}
|
Why do user-defined string literals and integer literals have different behavior?
By : Sereen Vinola
Date : March 29 2020, 07:55 AM
Hope that helps That's how floating point literals work!! Add a pair of parentheses and it should work: code :
std::cout << (4_s).count();
std::cout << 4_s .count();
// ^ Space here!
long double operator""_E(long double);
long double operator""_a(long double);
int operator""_p(unsigned long long);
auto x = 1.0_E+2.0; // error
auto y = 1.0_a+2.0; // OK
auto z = 1.0_E +2.0; // OK
auto w = 1_p+2; // error
auto u = 1_p +2; // OK
test.cpp:19:10: error: unable to find numeric literal
operator 'operator""_E+2.0'
^^^^^^
auto x = 1.0_E+2.0; // error
|
» 
code :
Related Posts :