#include <stdio.h>
#include <conio.h>
void main ()
{
clrscr () ;
char a [5];
puts ("K?");
gets (a);
fflush (stdin);
if (a = ("K"))
{
puts (a);
}
else
{
puts ("BAE");
}
getch ();
}
The 10th line shows an Lvalue error while compiling, please help. this is my first program ever and this is my first day ever on coding and I'm self teaching.
fflush(stdin) is Undefined Behavior
the comparison operator in C is a ==; = is the assignment operator. You can't assign anything to an array, so an error is thrown
you can compare strings using strcmp() from <string.h>; == just compares their addresses
I think that instead of the assignment in the if statement
if (a = ("K"))
you mean a comparison.
Nevertheless arrays do not have the comparison operator. You have to compare arrays element by element yourself. For character arrays there is function strcmp declared in header <string.h>.
Thus the valid code can look like
Also you should not use fflush function with standard input. Otherwise the program has undefined behaviour,
#include <string.h>
//...
if ( strcmp( a, "K" ) == 0 )
//...
There are two mistakes in this line
if (a = ("K"))
First, you are using = where you meant ==. But that is wrong anyway, you cannot test for string equality in C. It has to be something like this, which tests the first character of the string and its terminator
if (a[0] == 'K' && a[1] == '\0')
or, you can use a library function
#include <string.h>
...
if (strcmp(a, "K") == 0) {
// are the same string.
}
Related
I am unable to find out if the ternary operator is present or not through following code.. Plz help.
#include <stdio.h>
#include <ctype.h>
#include <string.h>
void one();
int main()
{
FILE *fp;
fp=fopen("C:/Users/HP/Documents/NetBeansProjects/CD2/1.txt","r");
char c;
void one()
{
char c;
while((c=fgetc(fp)!=EOF))
{
if(c==':')
{
printf("\nThe ternary operator present");
return;
}
}
}
while((c=fgetc(fp))!=EOF)
{
printf("\n-->%c",c);
if(c=='?')
{
one();
}
}
return 0;
}
I want to know why this code doesn't work and say if the ternary operator is present or not in file 1.txt
The output shows all characters till '?' if we print them, but why it's not finding for a colon ':' ?
The exit condition of the while loop may be the problem. = has lesser precedence than != operator. So
(c=fgetc(fp)!=EOF)
gets evaluated like
(c= (fgetc(fp)!=EOF) )
See this for the precedence of various operators in C.
You could do
while((c=fgetc(fp))!=EOF)
instead. First assign the return value of fgetc() to c and then do the comparison.
ie, the variable c gets the result of the comparison which means the value is either 0 or 1.
Your program should be checking for the ? and the other operands as well.
See this.
A simple check may not be enough for cases like the 'operator' being part of a string like
char str[]="the ternary operator is ?:";
Checking for the occurrence in such cases is a bit complex.
Edit:
As Jens pointed out, fgetc() returns an int not a char. So make c an int instead of a char.
See this post.
Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 5 years ago.
Improve this question
I'm a new person who loves to play around with coding. Recently I was going through a course on edx, and one of the exercises I need to complete has this small code snippet that keeps on giving Segmentation fault. I have taken out the faulty bit (everything else compiles nicely)
#include <stdio.h>
#include <string.h>
#include <cs50.h>
#include <ctype.h>
#include <stdlib.h>
int main (int argc, string argv[])
{
if (argc == 2 && isalpha(argv[1]))
{
int a = 0;
while (argv[1][a] == '\0')
{
a++;
printf("%c\n", argv[1][a]);
}
}
else
{
printf("Usage: ./programname 1-alphabetical word\n");
return 1;
}
}
The problem seems to be here: argv[1][a] but I can't for the life of me find out what, and how to fix it.
(1) isalpha(argv[1]) is wrong. This function expects a single character, but you are passing a pointer-to-string. That will certainly not give you any kind of expected result, and it's probably Undefined Behaviour into the bargain. You either need to loop and check each character, use a more high-level library function to check the entire string, or - as a quick and probably sense-changing fix - just check the 1st character as BLUEPIXY suggested: isalpha( argv[1][0] ) or isalpha( *argv[0] ).
(2) Your while condition is wrong. You are telling it to loop while the current character is NUL. This will do nothing for non-empty strings and hit the next problem #3 for an empty one. You presumably meant while (argv[1][a] != '\0'), i.e. to loop only until a NUL byte is reached.
(3) You increment index a before trying to printf() it. This will index out of range right now, if the input string is empty, as the body executes and then you immediately index beyond the terminating NUL. Even if the loop condition was fixed, you would miss the 1st character and then print the terminating NUL, neither of which make sense. You should only increment a once you have verified it is in-range and done what you need to do with it. So, printf() it, then increment it.
2 and 3 seem most easily soluble by using a for loop instead of manually splitting up the initialisation, testing, and incrementing of the loop variable. You should also use the correct type for indexing: in case you wanted to print a string with millions or billions of characters, an int is not wide enough, and it's just not good style. So:
#include <stddef.h> /* size_t */
for (size_t a = 0; argv[1][a] != '\0'; ++a) {
printf("%c\n", argv[1][a]);
}
isalpha(argv[1]) looks incorrect and should probably be isalpha(argv[1][0])
isalpha takes a character but you entered in a string to the function
another thing that sticks out as wrong is argv[1][a] == '\0' the ==
should be != this will mean that the while loop will stop once it hits the \0
perhaps
if (argc == 2)
{
int a = 0;
while (argv[1][a] != '\0')
{
if (isalpha(argv[1][a])
printf("%c\n", argv[1][a]);
a++;
}
}
may be what you are looking for?
The only reason for the segmentation fault I see is this subexpression of the if statement
if (argc == 2 && isalpha(argv[1]))
^^^^^^^^^^^^^^^^
There is specified an argument of an incorrect type. The expression argv[1] has the type char * while the function requires an object of character type that is interpreted as unsigned char and promoted to the type int.
So when the promoted argument has a negative value (except the EOF value) the function isalpha has undefined behavior.
From the C Standard (7.4 Character handling <ctype.h>)
1 The header <ctype.h> declares several functions useful for
classifying and mapping characters.198) In all cases the argument is
an int, the value of which shall be representable as an unsigned
char or shall equal the value of the macro EOF. If the argument has
any other value, the behavior is undefined.
You should write the if statement either like
if (argc == 2 && isalpha( ( unsigned char )argv[1][0] ) )
or like
if (argc == 2 && isalpha( ( unsigned char )*argv[1] ) )
Also there is a bug in the while statement that will execute never if the argument is not an empty string. I think you mean the following
int a = 0;
while ( argv[1][a] != '\0' )
{
printf("%c\n", argv[1][a]);
a++;
}
or for example like
int a = 0;
while ( argv[1][a] )
{
printf("%c\n", argv[1][a++]);
}
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;
}
Could someone point me to the error in this code. Iam using codingblocks IDE.
#include <stdio.h>
main()
{
char a[100],b[100];int i,j=0;
scanf("%s",&a);
for(i=strlen(a)-1,j=0;i>=0;i--)
{
b[j]=a[i];
j=j+1;
}
b[j]='\0';
if(a==b) # on printing i get both 'a' and 'b' as equal however this condition still remains
# false
printf("true"); #doesnot print?
}
Change this statement
if(a==b)
to
if ( strcmp( a, b ) == 0 )
Otherwise you are comparing addresses of the first elements of the arrays.
Also you need to include header <string.h>. Function main shall have return type int. Change this statement
scanf("%s",&a);
to
scanf( "%s", a);
Take into account that there is no need to define second array that to determine whether a string is a palindrome. You could do this check "in place".
Few issues in your code
You never copy anything in b, so that array has random characters.
a==b will always be false because they are character arrays not pointers and both contain different values.
If you are reading string, you don't need & for char array, so the scanf() should be scanf("%s",a);
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.