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");
Related
This question already has answers here:
How do I properly compare strings in C?
(10 answers)
Closed 5 years ago.
char str[6];
do
{
printf("Enter the string you wanna check:");
scanf("%s", str);
}
while(str != "exit");
Why does this not work?
str will never equal "exit", because you're comparing the addresses of two different sections of memory. You probably want to compare the contents of the strings, for which there is a function strcmp().
"exit" is a char[5] generated by the compiler at some address in the data segment. This address is definitely different from the address of str, as two different objects cannot occupy the same location in memory.
The != operator between expressions of type char[] compares two pointers. These two pointers are the address of "exit" and the address of str, which, as I have already explained, will never be equal.
So, the expression str != "exit" will never evaluate to true. Which brings us to another point: your compiler should have issued a warning about this condition being always false. Which means that you are trying to program without -Wall. Don't do this, you are never going to get very far. Always use the highest warning level, and when you see warnings, always fix them.
To correct the problem, do as user3121023 suggested in a comment, and use strcmp() to compare strings.
The short answer is: it does not work because you must use strcmp(str, "exit") to compare the strings and loop for as long as the return value of strcmp() is not 0.
The long answer is: there are more problems in this little code fragment:
The array into which you read a word is very short and you do not limit the number of characters scanf() is likely to store there. Any user input longer than 5 non space characters will cause undefined behavior.
You do not check the return value of scanf(). A premature end of file, such as redirecting input from an empty file, will cause an infinite loop.
Here is how the code can be written in a safer way:
#include <stdio.h>
int main(void) {
char str[80];
for (;;) {
printf("Enter the string you wanna check:");
if (scanf("%79s", str) != 1 || strcmp(str, "exit") == 0)
break;
}
return 0;
}
As suggested above, use strcmp from the <string.h> header file.
char str[6];
do {
printf("Enter the string you wanna check:");
scanf("%s", str);
} while(!strcmp(str, "exit"));
Try :
#include <stdio.h>
#include <string.h>
int main() {
char str[6];
do
{
printf("Enter the string you wanna check:");
scanf("%s", str);
}
while(strcmp(str, "exit") != 0);
return 0;
}
I have this code sample. There is a scanf to hold the input String values from keyboard (i.e Lotus). But even if I type the word Lotus correctly It does not execute the relevant if statement. **Is there any problem with my scanf function???
#include<stdio.h>
int main()
{
char landType,houseType[100],wantToContinue,wantToContinue2;
float installment;
wantToContinue='Y';
while(wantToContinue == 'Y' || wantToContinue == 'y') {
printf("Land Type : ");
scanf(" %c",&landType);
if(landType == 'A') {
printf("Type of House: ");
scanf(" %s", houseType);
if(houseType == "Lotus") {
//this won't go inside if statement even if I type Lotus correctly
installment=4000000-500000;
printf("Monthly Installment : R.s %.2f\n",installment/120);
printf("Do you want to Continue?(Y/y or N/n) : ");
scanf(" %c",&wantToContinue2);
if(wantToContinue2 == 'Y' || wantToContinue2 == 'y') {
wantToContinue=wantToContinue2;
printf("\n");
}else{
wantToContinue='N';
}
}
}
}
}
Be careful when comparing two strings in C. You should use the strcmp function from the string.h library, like so:
if(strcmp("Lotus", houseType) == 0)
When you write if(houseType=="Lotus") you are actually comparing the base address of the two strings and not their actual content.
In C, strings cannot be compared using ==. This is because strings are not a basic data type in C, that is C does not inherently understand how to compare them - you must use a function instead.
A standard function for comparing strings in C is strcmp(), e.g.:
if (strcmp(houseType, "Lotus") == 0)
{
// do some work if the strings are equal
}
To explain further, your initial attempt to compare strings using housetype == "Lotus" actually compares the address where the first character of the character array houseType is stored with the address where the first character of the character array "Lotus" is stored.
This happens because strings in C are just arrays of characters - they are not an inherent data type, C does not, as such, understand the difference between an array of integers and an array of characters, they are all just numbers arranged contiguously somewhere in memory and it treats them as such unless you specifically use code that operates on them as strings instead.
Why is this crashing when I input the string? I don't think I'm reading in the string right but the program gives me an error on the first 'scanf.' The program should be correct but this is C not C++. Most help that I could find was for C++.
//Andrei Shulgach
//April 27th, 2015
/*A string is a palindrome if it can be read forward and backward with the same
meaning. Capitalizations and spacing are ignored.*/
#include <stdio.h>
#include <stdlib.h>
int newStrCmp (const char *string1, const char *string2);
int main()
{
//Local Declarations
int dummy, value;
char string1[100],string2[100];
printf("Please enter the 1st string: ");
scanf_s("%99s", string1[100]);
printf("\nPlease enter the 2nd string: ");
scanf_s("%99s", string2[100]);
//Call Function and get value
value = newStrCmp(string1, string2);
if (value == 0)
printf("The strings are equal.\n");
else
printf("The strings are not equal.\n");
scanf_s("%d",&dummy);//Keep Window Open
return 0;
}
int newStrCmp (const char *string1, const char *string2)
{
//Local Declarations
int value = 0;
while (string1[value] == string2[value])
{
if (string1[value] == '\0' || string2[value] == '\0')
break;
value++;
}
if (string1[value] == '\0' && string2[value] == '\0')
return 0;
else
return -1;
}
You must enable all compiler warnings that you can get from your compiler; the above code shouldn't have compiled.
This:
scanf_s("%99s", string1[100]);
invokes undefined behavior since it indexes outside the 100-character string1 array. Remember that C arrays are indexed from 0. It also fails to comply with scanf_s()'s requirement that the size be specified for all string conversions.
It then probably1 causes more undefined behavior, when scanf_f() interprets a single character as a buffer address where input is to be stored (assuming the call happens, of course).
This is not valid code.
It should simply pass the address of the first character in the array:
scanf_s("%99s", string1, sizeof string1);
Here, string1 is the same as &string1[0]; the name of an array evaluates to the address of its first element in many contexts. We then use sizeof string1 as the third argument to specify the size of the string1 buffer, which is required.
1 You cannot reason about what happens after undefined behavior has happened with any certainty.
The issue is you are 'scanf'ing into index 100 of your length 100 buffer, i.e. out-of-bounds.
scanf_s("%99s", string1);
Also your comparison would be safer as a for loop that ensures value is less than 100, rather than a while loop.
In the future, please include the error message.
So basically I wanted to create a program in C, in wich you would input 2 character long string (mix of letter and noumber ex.r1,u2,i3,i4,r6) to be the input in my program. Later I want to put this string in SWITCH. Is this possible?
Here's my simple sourcecode. Please correct me on any mistakes :)
#include <stdio.h>
int main(void)
{
char string[2];
scanf("%s", &string);
switch (string)
{
case 'u1' :printf("%s\n", string);break;
default :printf("ERROR");break;
}
return 0;
}
Create a code based on the string and switch on that.
#define Code(a,b) (a + 256*b)
char string[3]; // 3 not 2
if (scanf("%2s", string) != 1) { // No &
Handle_Error();
}
int scode = Code(string[0], string[1]);
switch (scode) {
case Code('u', '1') : printf("%s\n", string); break;
case Code('r', '2') : printf("r2\n"); break;
...
default :printf("ERROR");break;
}
A switch(x) needs an integer value for x and string is an array. So the original approach will not work.
The program can use an integer based on the string for x and use the same method for generating the case values. Since there is only the first 2 char of the string are of interest, the int value is unique.
No, this is not possible. Switch only works with integral types in C (int, short, long, etc, as well as types defined with enum).
You can however use a simple if-else construct to get the same behavior:
if (strcmp(string, "ui" ) == 0) //test for string equality
{
printf("%s\n", string);
}
else
{
printf("ERROR")
}
We use strcmp instead of == because we are dealing pointers which almost certainly not compare equal even when the two strings have the same content.
strcmp(str1, str2) == 0 is the standard idoim in C for comparing two strings.
strcmp returns an integer representing how two strings compare to each other. 0 means they are equal, a negative number means that the first string is lexographically "less than" the second, and a positive number means that the first string is lexographically "greater than" the second. More info can be found here.
A switch won't work here.
You need to use an if/else if construct and strcmp to compare the strings.
Also, you need at least 3 characters in your input array so that it can hold the two input characters and the terminating null character.
Of course, such a small buffer can easily overflow.
I am working on learning C and am using some practice problems from the Python book I recently finished. My C book is in the mail, but I wanted to get a head start. I was putting together a simple temperature conversion program and for some reason it always jumps to the 'Else' clause in my conditional... I'm sure I'm missing something simple, but I can't seem to figure it out. Any ideas?:
#include<stdio.h>
main()
{
float temp_c, temp_f;
char convert_from[1];
printf("Convert from (c or f): ");
scanf("%c", &convert_from);
if(convert_from == "c")
{
printf("Enter temperature in Celsius: ");
scanf("%f", &temp_c);
temp_f=(1.8*temp_c)+32;
printf("The temperature in Fahreinheit is: %f \n", temp_f);
}
else if(convert_from == "f")
{
printf("Enter temperature in Fahreinheit: ");
scanf("%f", &temp_f);
temp_c=(temp_f/1.8)-32;
printf("The temperature in Celsius is: %f \n", temp_c);
}
else
printf("Invalid choice. \n");
}
If you are comparing characters do this:
char convert_from;
printf("Convert from (c or f): ");
scanf("%c", &convert_from);
if (convert_from == 'c')
{
Otherwise you can't perform the comparison with a string literal "c" (note the double quotes) like that.
In the expression:
if (convert_from == "c")
convert_from array is converted to a pointer to char, so you are basically comparing a pointer to char to another pointer to char. "c" is a string literal while 'c' is a char (note the use of "" in the first case and of '' in the second case).
With a char convert_from[1]; declaration, here would be the correct code:
char convert_from[1];
scanf("%c", convert_from);
if (convert_from[0] == 'c')
but it is more natural to directly use a char instead of an array 1 of char:
char convert_from;
scanf("%c", &convert_from);
if (convert_from == 'c')
Well two problems here.
First: You want only to read a single character, so declaring
char convert_from;
instead of an array of chars with size 1 (as char convert_from[1] did).
And secondly you need to actually compare to a single character, so you would need to do
if (convert_from == 'c') ...
instead of "c", because "foo" is a string in C which in turn is a pointer to a constant
character array (const char *).
Additionally: which compiler did you use? Mine (llvm-gcc 4.2) warned me about the issues. So either your compiler is quite bogus or you will urgently have to pay attention to compiler warnings. This might be difficult, as warnings are no errors, however, warnings are there for a reason :-)
In C, you cannot compare strings using == (what happens if you do is that the memory locations of the strings get compared, which will yield different results in most cases).
Also scanf("%c", &convert_from); is wrong. The array itself will already decay into a pointer, so scanf("%c", convert_form); will suffice. In this case however, convert_form will not contain something your C library will consider a string (strings are null-terminated in C). A minimally invasive change for getting your code to work would be changing
if (convert_from == "f") [...]
to
if (covert_form[0] == 'f') [...]
(mind the '' instead of the "", which is a character literal, which basically is just a number and can thus be compared using ==).
A more idiomatic way to do this would be to declare convert_form as char convert_form and then use scanf("%c", &convert_form);, which would accomplish the same as the above.
First of all since you are reading only one character at a time
define it as
char convert_from;
next it is not advisable to compare strings directly
hence the statement should be
if(convert_from == 'c')