For C++, use std::string and compare using string::compare. For C use strcmp. If your (i meant your programs) strings (for some weird reason) aren’t nul terminated, use strncmp instead. But why would someone not use something as simple as == for std::string ?
In C++, use std::string, and the operator+, it is designed specifically to solve problems like this. #include <iostream> #include <string> using namespace std; int main() { string foo( “hello” ); string test( “how are” ); cout << foo + ” , ” + test; return 0; }
The simplest way is to use the conversion macros: CW2A CA2W etc… MSDN
#include<string.h> #include<stdio.h> int main() { char input[16] = “abc,d”; char *p; p = strtok(input, “,”); if(p) { printf(“%s\n”, p); } p = strtok(NULL, “,”); if(p) printf(“%s\n”, p); return 0; } you can look this program .First you should use the strtok(input, “,”).input is the string you want to spilt.Then you use the strtok(NULL, “,”). If … Read more
The character ‘\’ is a special character and needs to be escaped when used as part of a String, e.g., “\”. Here is an example of a string comparison using the ‘\’ character: if (invName.substring(j,k).equals(“\\”)) {…} You can also perform direct character comparisons using logic similar to the following: if (invName.charAt(j) == ‘\\’) {…}
You could use memcpy. If the multidimensional array size is given at compile time, i.e mytype myarray[1][2], then only a single memcpy call is needed memcpy(dest, src, sizeof (mytype) * rows * columns); If, like you indicated the array is dynamically allocated, you will need to know the size of both of the dimensions as … Read more
Use pointer subtraction. char *str = “sdfadabcGGGGGGGGG”; char *result = strstr(str, “abc”); int position = result – str; int substringLength = strlen(str) – position;
This has nothing to do with structs – arrays in C are not assignable: char a[20]; a = “foo”; // error you need to use strcpy: strcpy( a, “foo” ); or in your code: strcpy( sara.first, “Sara” );
Try this: char letter=”c”; if (letter == ‘z’) nextChar=”a”; else if (letter == ‘Z’) nextChar=”A”; else nextChar = (char)(((int)letter) + 1); This way you have no trouble when the char is the last of the alphabet.
Try this: ORDER BY CAST(thecolumn AS int)