I'm a noob at C programming and I'm having some difficulties making a string list and searching for a specific element.
#include <stdio.h>
#include <string.h>
# define MAX 6
int main(){
char word[MAX];
char x[MAX][20];
int i;
strcpy(x[0], "One");
strcpy(x[1], "Two");
strcpy(x[2], "Three");
strcpy(x[3], "Four");
strcpy(x[4], "Five");
strcpy(x[5], "Six");
printf("%s", "Search:");
scanf("%s", word);
for (i=0; i<6; i++) {
if (x[i] == word) {
printf("%s", "Found a match!");
}
}
return 0;
}
It's never executing the statement present in the if block (i.e, printf("Found a match!")) . Any idea why it is not executing the above mentioned statement?
Thanks!
Use
if(strcmp(x[i],word) == 0)
printf("Found match\n");
== can't be used to compare strings as you are doing it.
This only compares the pointers and not the strings
It never returns "Found a match!". Any idea why?
Reason:
In C, array names are converted to pointers to their first elements ( with some exceptions there). x[i] == word is comparing two pointers instead of comparing strings. Since the base addresses of both arrays are different, comparison returns a false value.
Correction:
Use strcmp to compare two strings.
This
if (x[i] == word)
should be
if (strcmp(x[i], word) == 0)
In c a predefined function is present in string.h library it is strcmp as stated by other users function int strcmp(const char *str1, const char *str2) compares the string pointed to bystr1 to the string pointed to by str2.u can write your own function for comparing strings and use it.
I want you to conceptually understand why we can't use == in c unlike c++ as c don't contain anything like string class(c is purely procedural) so that u can create object of it and use it.hence c uses char array to represent a string .if u examine ur code x[i] == word compares starting addresses of char arrays/strings x[i],word. I believe u understood the concept . now I want to explain that u can use pointers here i.e
if (*x[i] == *word)
printf("Found a match!");
Works fine as u can understand that here we are comparing two strings directly by pointing to their address locations.sorry if I have provided unwanted info due to my inexperience in SO as this my first answer in SO.
use strcmp() function for comparing two string. when two string is match its result is 0.
so you can change like :
if ( ! strcmp(word,x[i]) ) // when match result will be `! 0 = 1`
printf("Found Match\n");
in the C programming language, the == operator is not working for comparing strings(as others wrote before me). I advice to try using a really nice feature in C++ called string. It is builded in the standard library, and with this, you can use the == operator for comparing.
#include <iostream>
using namespace std;
int main(void)
{
string a = "Apple";
if( a == "Apple" ) cout << "This is working" << endl;
}
Related
I'm a rookie programmer trying to run a simple code on VS code.
#include<stdio.h>
int main()
{
char* a;
printf("Enter a char");
scanf("%s",&a);
if (a = "yes")
{
printf("Number is 30");
}
else if (a = "no")
{
printf("Number is 50");
}
else{
printf("oops");
}
return 0;
}
I guess looking at the code you guys can figure out what I'm trying to do, if the user enters "yes", a specific sentence need to be displayed and similarly for "no".
The problem here is whatever I write in the input, it will always print the first statement, "Number is 30". I've tried running similar codes but ended up with the same output.
If possible, please explain me how to use char,strings,arrays with if-else statements.
There are several misunderstandings in the posted code.
First there is a misunderstanding of char versus string. A char is for instance a single letter, a single special character like ., ;, etc. (see note1) while a string is a serie of chars. So
'y' is a char
"yes" is a string
You print "Enter a char" but from the code it's obvious that you really want "Enter a string".
This leads to the next problem. To input a string using scanf you need to pass a "pointer to char". Your code pass "a pointer to pointer to char" due to the &. Further the passed pointer must point to some memory. So you need:
char a[10]; // Make it an array of char so that it can hold a string
printf("Enter a string, max 9 characters");
scanf("%9s", a); // No & before a and width specifier used to avoid buffer overflow
Now this part
if (a = "yes")
is not the way to compare two strings in C. For that you need the function strcmp - like:
if (strcmp(a, "yes") == 0)
Putting it together it's like:
int main()
{
char a[10];
printf("Enter a string, max 9 characters");
scanf("%9s", a);
if (strcmp(a, "yes") == 0){
printf("Number is 30");
}
else if (strcmp(a, "no") == 0)
{
printf("Number is 50");
}
else
{
printf("oops");
}
return 0;
}
That said, I don't understand why you print stuff like: "Number is 30" but that's kind of irrelevant here.
note1: The type char is actually an integer type, i.e. a number, but the common use is to map these numbers to characters using ASCII encoding.
There are different ways to initialize a variable to access C string.
char *char_ptr = "Hello";
This initializes char_ptr to point to the first character of the read-only string "Look Here".A C string initialized through a character pointer cannot be modified. When a C string is initialized this way, trying to modify any character pointed to by char_ptr is undefined behaviour. An undefined behaviour means that when a compiler encounters anything that triggers undefined behaviour, it is allowed to do anything it seems appropriate.
A more convenient way to define strings that can be modified is to use:
char str[];
This way you can modify any character in the C string
p.s you also need to use strcmp() for the if statement
You can take string input in C using
scanf(“%s”, str);
And to compare the string you need to use:
strcmp(str1, "yes");
For example,
Supposing that I've created a string called string[100] and assigned some values to it, how can I know if the value of string[n] is "m"?
I've thought about creating another char variable and assigning "m" to it and then comparing string[n] to char with strcmp, but it seems kinda like a waste of code creating a variables just for that since on the program I'm trying to code I'll have to compare string[n] with a lot of stuff.
Thanks in advance and sorry for my poor knowledge on C, i'm starting to learn it today.
You're just beginning, and strings in C are not as "user-friendly" as they are in some other programming languages, so there's nothing wrong with asking.
In C a string is just an array of characters, with a trailing null character ('\0') to indicate the end of the string. Each character in that array, though, as mentioned in the comments, is just a character, which is a small integer that can be compared with simple operators like ==, !=, >, <, etc.
If we want to know if a string begins with an 'm' for instance, we can write a simple function that checks the first element of the array. Please note that C arrays are indexed starting at 0.
int starts_with(char ch, char *str) {
return s[0] == ch;
}
Once you learn how to use loops and other control flow mechanisms in C, you'll be able to do more complex operations on strings. As an example, determining if a character occurs in a string.
int contains_char(char ch, char *str) {
for (char *c = str; *c; c++) {
if (c == ch) {
return 1;
}
}
return 0;
}
How this comparison works?
#include <stdio.h>
#include <string.h>
int main(void) {
char arr[]="WELCOME";
char arr1[]="WELCOME";
if (arr<=arr1)
printf("equal");
else
printf("not equal");
return 0;
}
In this program that if condition this always go to else,
so please some one help me how that comparison done here.
The comparison ends up asking where the two arrays arr and arr1 are located with respect to each other in memory. It is asking which one has the lower address.
If you're trying to compare the string values contained in the two arrays, use strcmp:
if(strcmp(arr, arr1) <= 0) ...
As far as my knowledge goes, arr or as a matter of fact any array variable stores the address of the first element in the array. So when you do arr<=arr1 you are actually comparing their addresses. So that's what it does.
P.S. you can print the addresses just in case you want to check. As in:
printf("\n%p", (void*) arr);
From the looks of your program, it seems that you want to compare the two strings. You then have to do:
if(strcmp(arr,arr1)==0)
printf("Entered strings are equal.\n");
else
printf("Entered strings are not equal.\n");
This question already has answers here:
Using the equality operator == to compare two strings for equality in C [duplicate]
(9 answers)
Closed 8 years ago.
In the following working code; instead of using *tofind, if I directly use the comparison
if(*argv[i] == "and")
it fails.
Why would that be?
/**
* Find index of the word "and"
* ./a.out alice and bob
*/
int main(int argc, char **argv) {
int i = 0;
char *tofind = "and";
while (argv[i] != NULL) {
if(*argv[i] == *tofind) {
printf("%d\n", i + 1);
break;
}
++i;
}
return 0;
}
if(*argv[i] == "and") shouldn't compile, I think you mean if (argv[i] == "and"), that would compare the pointers of the two, not the string content.
if (*argv[i] == *tofind) doesn't work as you expected either, it only compares the first character.
To compare strings, use strcmp():
if (strcmp(argv[i], tofind) == 0)
A "char*" is officially a pointer to a single char, for example to find points to a letter 'a'. You know that there are two more chars and a nul char, but officially it points to a single char.
Therefore, *argv[i] is the first character of an argument, and *tofind is always the letter 'a', so your code checks if the first character of an argument is an 'a'. Look at the strcmp function, which compares whole strings.
look at the type of
*argv[i] //its type is char
and of "and"
"and" //its type is const char * as it is decayed into pointer
so thats why you are unable to compare them.
while the type of
*tofind
is char and you can now compare the two.for more details see FAQs section 6.
This question already has answers here:
C Strings Comparison with Equal Sign
(5 answers)
Closed 9 years ago.
i've written some simple code as an SSCCE, I'm trying to check if string entered is equal to a string i've defined in a char pointer array, so it should point to the string and give me a result. I'm not getting any warnings or errors but I'm just not getting any result (either "true" or "false")
is there something else being scanned with the scanf? a termination symbol or something? i'm just not able to get it to print out either true or false
code:
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#define LENGTH 20
//typedef char boolean;
int main(void)
{
const char *temp[1];
temp[0] = "true\0";
temp[1] = "false\0";
char var[LENGTH];
printf("Enter either true or false.\n");
scanf("%s", var);
if(var == temp[0]) //compare contents of array
{
printf("\ntrue\n");
}
else if(var == temp[1]) //compare contents of array
{
printf("\nfalse\n");
}
}
const char *temp[1];
This defines tmp an array that can store 1 char* element.
temp[0] = "true\0";
Assigngs to the first element. This is okay.
temp[1] = "false\0";
Would assign to the second element, but temp can only store one. C doesn't check array boundaries for you.
Also not that you don't have to specify the terminating '\0' explicitly in string literals, so just "true" and "false" are sufficient.
if(var == temp[0])
This compares only the pointer values ("where the strings are stored"), not the contents. You need the strcmp() function (and read carefully, the returned value for equal strings might not be what you expect it to be).
Use strcmp for comparing strings:
#include <stdio.h>
int main(void)
{
const int LENGTH = 20;
char str[LENGTH];
printf("Type \"true\" or \"false\:\n");
if (scanf("%19s", str) != 1) {
printf("scanf failed.");
return -1;
}
if(strcmp(str, "true") == 0) {
printf("\"true\" has been typed.\n");
}
else if(strcmp(str, "false") == 0) {
printf("\"false\" has been typed.\n");
}
return 0;
}
and also note that:
string literals automatically contain null-terminating character ("true", not "true\0")
const int LENGTH is better than #define LENGTH since type safety comes with it
"%19s" ensures that no more than 19 characters (+ \0) will be stored in str
typedef char boolean; is not a good idea
unlikely, but still: scanf doesn't have to succeed
and there is no #include <iostream> in c :)
== checks for equality. Let's see what you're comparing.
The variable var is declared as a character array, so the expression var is really equivalent to &var[0] (the address of the first character in the var array).
Similarly, temp[0] is equivalent to &temp[0][0] (the address of the first character in the temp[0] array).
These addresses are obviously different (otherwise writing var would automatically write temp[0] as well), so == will always return 0 for your case.
strcmp, on the other hand, does not check for equality of its inputs, but for character-by-character equality of the arrays pointed to by its inputs (that is, it compares their members, not their addresses) and so you can use that for strings in C. It's worth noting that strcmp returns 0 (false) if the strings are equal.