Comparing chars stored in 2d arrays c++/c [duplicate] - c++

Question

I am trying to get a program to let a user enter a word or character, store it, and then print it until the user types it again, exiting the program. My code looks like this:
#include <stdio.h>
int main()
{
char input[40];
char check[40];
int i=0;
printf("Hello!\nPlease enter a word or character:\n");
gets(input); /* obsolete function: do not use!! */
printf("I will now repeat this until you type it back to me.\n");
while (check != input)
{
printf("%s\n", input);
gets(check); /* obsolete function: do not use!! */
}
printf("Good bye!");
return 0;
}
The problem is that I keep getting the printing of the input string, even when the input by the user (check) matches the original (input). Am I comparing the two incorrectly?

Answer 1

You can't (usefully) compare strings using != or ==, you need to use strcmp:
while (strcmp(check,input) != 0)
The reason for this is because != and == will only compare the base addresses of those strings. Not the contents of the strings themselves.

Answer 2

Ok a few things: gets is unsafe and should be replaced with fgets(input, sizeof(input), stdin) so that you don't get a buffer overflow.
Next, to compare strings, you must use strcmp, where a return value of 0 indicates that the two strings match. Using the equality operators (ie. !=) compares the address of the two strings, as opposed to the individual chars inside them.
And also note that, while in this example it won't cause a problem, fgets stores the newline character, '\n' in the buffers also; gets() does not. If you compared the user input from fgets() to a string literal such as "abc" it would never match (unless the buffer was too small so that the '\n' wouldn't fit in it).

Answer 3

Use strcmp.
This is in string.h library, and is very popular. strcmp return 0 if the strings are equal. See this for an better explanation of what strcmp returns.
Basically, you have to do:
while (strcmp(check,input) != 0)
or
while (!strcmp(check,input))
or
while (strcmp(check,input))
You can check this, a tutorial on strcmp.

Answer 4

You can't compare arrays directly like this
array1==array2
You should compare them char-by-char; for this you can use a function and return a boolean (True:1, False:0) value. Then you can use it in the test condition of the while loop.
Try this:
#include <stdio.h>
int checker(char input[],char check[]);
int main()
{
char input[40];
char check[40];
int i=0;
printf("Hello!\nPlease enter a word or character:\n");
scanf("%s",input);
printf("I will now repeat this until you type it back to me.\n");
scanf("%s",check);
while (!checker(input,check))
{
printf("%s\n", input);
scanf("%s",check);
}
printf("Good bye!");
return 0;
}
int checker(char input[],char check[])
{
int i,result=1;
for(i=0; input[i]!='\0' || check[i]!='\0'; i++) {
if(input[i] != check[i]) {
result=0;
break;
}
}
return result;
}

Answer 5

Welcome to the concept of the pointer. Generations of beginning programmers have found the concept elusive, but if you wish to grow into a competent programmer, you must eventually master this concept — and moreover, you are already asking the right question. That's good.
Is it clear to you what an address is? See this diagram:
---------- ----------
| 0x4000 | | 0x4004 |
| 1 | | 7 |
---------- ----------
In the diagram, the integer 1 is stored in memory at address 0x4000. Why at an address? Because memory is large and can store many integers, just as a city is large and can house many families. Each integer is stored at a memory location, as each family resides in a house. Each memory location is identified by an address, as each house is identified by an address.
The two boxes in the diagram represent two distinct memory locations. You can think of them as if they were houses. The integer 1 resides in the memory location at address 0x4000 (think, "4000 Elm St."). The integer 7 resides in the memory location at address 0x4004 (think, "4004 Elm St.").
You thought that your program was comparing the 1 to the 7, but it wasn't. It was comparing the 0x4000 to the 0x4004. So what happens when you have this situation?
---------- ----------
| 0x4000 | | 0x4004 |
| 1 | | 1 |
---------- ----------
The two integers are the same but the addresses differ. Your program compares the addresses.

Answer 6

Whenever you are trying to compare the strings, compare them with respect to each character. For this you can use built in string function called strcmp(input1,input2); and you should use the header file called #include<string.h>
Try this code:
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main()
{
char s[]="STACKOVERFLOW";
char s1[200];
printf("Enter the string to be checked\n");//enter the input string
scanf("%s",s1);
if(strcmp(s,s1)==0)//compare both the strings
{
printf("Both the Strings match\n");
}
else
{
printf("Entered String does not match\n");
}
system("pause");
}

Answer 7

You need to use strcmp() and you need to #include <string.h>
The != and == operators only compare the base addresses of those strings. Not the contents of the strings
while (strcmp(check, input))
Example code:
#include <stdio.h>
#include <string.h>
int main()
{
char input[40];
char check[40] = "end\n"; //dont forget to check for \n
while ( strcmp(check, input) ) //strcmp returns 0 if equal
{
printf("Please enter a name: \n");
fgets(input, sizeof(input), stdin);
printf("My name is: %s\n", input);
}
printf("Good bye!");
return 0;
}
Note1: gets() is unsafe. Use fgets() instead
Note2: When using fgets() you need to check for '\n' new line charecter too

Answer 8

You can:
Use strcmp() from string.h, which is the easier version
Or if you want to roll your own, you can use something like this:
int strcmp(char *s1, char *s2)
{
int i;
while(s1[i] != '\0' && s2[i] != '\0')
{
if(s1[i] != s2[i])
{
return 1;
}
i++;
}
return 0;
}
I'd use strcmp() in a way like this:
while(strcmp(check, input))
{
// code here
}

Answer 9

How do I properly compare strings?
char input[40];
char check[40];
strcpy(input, "Hello"); // input assigned somehow
strcpy(check, "Hello"); // check assigned somehow
// insufficient
while (check != input)
// good
while (strcmp(check, input) != 0)
// or
while (strcmp(check, input))
Let us dig deeper to see why check != input is not sufficient.
In C, string is a standard library specification.
A string is a contiguous sequence of characters terminated by and including the first null character.
C11 §7.1.1 1
input above is not a string. input is array 40 of char.
The contents of input can become a string.
In most cases, when an array is used in an expression, it is converted to the address of its 1st element.
The below converts check and input to their respective addresses of the first element, then those addresses are compared.
check != input // Compare addresses, not the contents of what addresses reference
To compare strings, we need to use those addresses and then look at the data they point to.
strcmp() does the job. §7.23.4.2
int strcmp(const char *s1, const char *s2);
The strcmp function compares the string pointed to by s1 to the string pointed to by s2.
The strcmp function returns an integer greater than, equal to, or less than zero,
accordingly as the string pointed to by s1 is greater than, equal to, or less than the string pointed to by s2.
Not only can code find if the strings are of the same data, but which one is greater/less when they differ.
The below is true when the string differ.
strcmp(check, input) != 0
For insight, see Creating my own strcmp() function

Answer 10

#include<stdio.h>
#include<string.h>
int main()
{
char s1[50],s2[50];
printf("Enter the character of strings: ");
gets(s1);
printf("\nEnter different character of string to repeat: \n");
while(strcmp(s1,s2))
{
printf("%s\n",s1);
gets(s2);
}
return 0;
}
This is very simple solution in which you will get your output as you want.