void main()
{
int x,y;
scanf("%d", &x); <-1
y=(x>5?3:4);
printf("%d",y); ->4
}
What is the use of: in y=(x>5?3:4); ?
It is known as ternary operator. This:
y= (x>5 ? 3 : 4);
means if x is greater than 5, to y will be assigned the value 3, otherwise (x is less or equal to 5) the value 4 will be assigned to y. That code is semantically the same as:
if(x > 5)
y = 3;
else
y = 4;
you can find a more in-depth explanation here.
"What is the use of : in y=(x>5?3:4); ?"
: represents the shorthand version of the else clause of an if-then-else expression when using the C ternary operator. So in your example:
y = (x>5 ? 3 : 4);
The English expression could be stated as:
if x is greater than 5, then set y equal to 3, else set
y equal to 4.
The traditional long version in C syntax would use the expression sequence:
if(x > 5)
{
//true path
y = 3;
}
else
{
//false path
y = 4;
}
It's a ternary operator or conditional operator.
Related
This question already has answers here:
Why is a condition like (0 < a < 5) always true?
(4 answers)
Closed 9 years ago.
I have this question & in the answer it says that due to left to right associativity the result would be 1 that is true for this statement. This is the code.
#include<stdio.h>
int main ()
{
int i=0,x=10,y=10,z=5;
i=x<y<z;
printf("\n\n%d",i);
return 0;
}
But x is greater than z here so how is this happening ?
The expression x
(x < y) < z
so it becomes
(10 < 10) < 5
which further is evaluated into
0 < 5
which is true.
I think you wanted something like this:
x < y && y < z
Because of operator precedence and associativity
i = x < y < z;
is parsed as:
i = ((x < y) < z);
After substituting the variable values, this becomes:
i = ((10 < 10) < 5);
Since 10 < 10 is false, this becomes:
i = (0 < 5);
Since 0 < 5 is true, that becomes:
i = 1;
x<y<z is not a single valid expression. Instead it evaluates x<y first (operator precedence is done left to right here) as true/false (false in this case as they're equal), converts it to an int value of 0, and then compares this value with z.
Use (x < y && y < z) instead.
It first evaluates x < y which is false (0), then 0 < z which is true (1).
WHat C compiler does is, in x<y<z;
starts from left, so as x is not less than y therefore it replaces that expression with '0'
so it becomes 0<z and as that is true. it set the variable to 1.
can any one help me understanding the output of this code?
#include <stdio.h>
int main()
{
int x = 1, y = 1;
for(; y; printf("%d %d \n", x, y))
{
y = x++ <= 5;
}
return 0;
}
the output is:
2 1
3 1
4 1
5 1
6 1
7 0
A for loop in the form of
for (a; b; c)
{
d;
}
is equivalent to
{
a;
while (b)
{
d;
c;
}
}
Now if we take your loop
for(; y; printf("%d %d \n", x, y))
{
y = x++ <= 5;
}
It is equivalent to
{
// Nothing
// Loop while y is non-zero
while (y)
{
// Check if x is less than or equal to 5, assign that result to y
// Then increase x by one
y = x++ <= 5;
printf("%d %d \n", x, y);
}
}
Now it should hopefully be easier to understand what's going on.
Also: Remember that for boolean results (like what you get as result from a comparison), true is equal to 1, and false is equal to 0.
The code is an obfuscated, ugly version of this:
#include <stdio.h>
#include <stdbool.h>
int main()
{
int x = 1;
bool y = true;
while(y == true)
{
y = (x++ <= 5);
printf("%d %d \n", x, (int)y);
}
return 0;
}
y in the original code serves as a boolean. Back in the ancient days, there were no boolean type in C so it was common to use int instead. The expression y = x++ <= 5; evaluates to 0 or 1, which is equivalent to false or true.
Note:
While the C language allows all manner of crazy stuff, you should never write for loops as in the original code. De facto standard is to write for loops like this:
The first clause of a for loop shall only contain iterator initialization.
The second clause should only contain the loop condition.
The third clause should only contain a change of the loop iterator, such as for example an increment (like i++).
A for loop that doesn't follow the above industry standard rules is badly written, no excuses.
The statement y = x++ <= 5; sets y to 1 if x is less than or equal to 5, or to 0 if x is greater than 5. It then increments x. The loop stops when y is equal to 0, that is, on the iteration after x was incremented above 5. Then it runs the printf() statement on each iteration.
You wouldn’t normally see a loop switch the loop condition and the loop body that way. This seems to be a pathological example of how it’s technically legal. You shouldn’t imitate it.
y is 1 (true) as long as x <= 5 and x increases every iteration.
Rewrite your code:
for (int x = 1, y = 1; y; y = x <= 5, x++) {
printf("%d %d\n", x, y);
}
or using while
int x = 1;
int y = 1;
while (y) {
printf("%d %d\n", x, y);
x++;
y = x <= 5;
}
The first component of the for loop is empty; which means values of variables haven't changed. The condition in the for loop is y. Which means that the loop will run till the condition y is true; which means till the value of y is anything other than 0.
When the value of y becomes equal to 0; the condition will become false and the loop will stop iterating.
Then comes the block execution...
y = x++ <= 5;
Here first the condition x <= 5 is checked. Note that x++ is post increment and so the value of x will be incremented after the execution of the statement. So, if the value of x was 1 before increment; it will check 1 <= 5 and not 2 <= 5 and after the execution of the statement; the value will become 2.
After that comes the third component of for loop which is the increment part. Here,
printf("%d %d \n", x, y)
simply prints x and y.
So, what happens is; when the loop starts, both x and y are 1.
The statement
y = x++ <= 5;
is encountered, thus the condition x++ <= 5 is checked, where x is 1. Since the condition is true; the value of y will be 1 and x will be incremented to 2. The same thing continues.... when x will be equal to 5, x++ <= 5 will check 5 <= 5 and the condition will be true and x will be incremented to 6. Now when it goes in the loop the next time; the condition will be false and y will instead be equal to 0. Thus, now when the loop checks the condition y, the condition will be false and thus the control will flow out of the loop and thus you see the result.
I built and dumped this code in IDA since I understand assembly better :D
This code is a bit obfuscated so here will be the normal version of it:
#include <stdio.h>
int x = 1;
int y = 1;
while(y != 0) {
if(x++ <= 5)
{
y = 1;
} else {
y = 0;
}
printf("%d %d\n",x,y);
}
that for loop in the source:
for(; y; printf("%d %d \n", x, y))
checks if y is a non-zero value and if it is, it will print the string you want. In the body of the loop, the result of the comparison (0 or 1) will be copied to y and the check continues.
Precedence of operator <= is higher than the operator =.
So, the statement:
y = x++ <= 5;
is equivalent to:
y = (x++ <= 5)
Because of post-increment ++ operator, the value of x returned first and then x is incremented.
The initial value of both x and yis 1.
Since y is 1, the for loop condition evaluates to true and the statement in the for loop block executed. But in the for loop of your code, in place of loop iterator update statement you have printf() statement which is printing the value of x and y after executing the loop body statements. The flow goes like this:
1 <= 5 evaluate to 1 which is assigned to y and after this x is 2,
Output : 2 1
2 <= 5 evaluate to 1 which is assigned to y and after this x is 3,
Output : 3 1
3 <= 5 evaluate to 1 which is assigned to y and after this x is 4,
Output : 4 1
4 <= 5 evaluate to 1 which is assigned to y and after this x is 5,
Output : 5 1
5 <= 5 evaluate to 1 which is assigned to y and after this x is 6,
Output : 6 1
6 <= 5 evaluate to 0 which is assigned to y and after this x is 7,
Output : 7 0
Now the value y is 0 and in your code, the for loop condition is y so the loop condition evaluates as false and loop exits.
I am a novice in C and having the following problem while compiling a simple code:
#include <stdio.h>
int main()
{
int i, a, b;
i = 3;
a = b = 0;
printf("Before: ");
printf("%d %d\n", a, b);
i == 3 ? a = 4 : a = 10; /* Line 9 */
printf("After: ");
printf("%d %d\n", a, b);
return 0;
}
Gives me error:
#gcc some.c
In function ‘main’:
some.c:9: error: lvalue required as left operand of assignment
I cannot understand it. What am i doing wrong?
This operator
i==3 ? a=4 : a = 10;
is equivalent to
( i==3 ? a=4 : a ) = 10;
Use instead
i==3 ? a=4 : ( a = 10 );
You need to assign outside the ternary operator, not inside of it, because the ternary operator binds tighter than assignment. So you should write
a = ((i == 3) ? 4 : 10);
What you wrote is equal to
(i == 3 ? a = 4 : a ) = 10;
In this code, an rvalue is being assigned to. This is the same error as writing
myfunc() = 16;
Cosmetic note: In this case it is better to use an if else statement instead of a conditional operator, because it is more clear here.
if (i == 3)
a = 4;
else
a = 10;
?: is a ternary operator.
ternary operator have higher precedence than '=' operator.
so 9th line of your code work like this..
(i == 3 ? a = 4 : a) = 10;
you have to use parenthesis or modify your sentence.
you can replace your 9th line with either of following line..
i == 3 ? a = 4 : (a = 10);
a = i == 3 ? 4 : 10;
a = (i == 3 ? 4 : 10);
Try this! :-)
Try this
if (i==3) a = 4; else a = 10;
The ternary operator does not work like this. You should use parenthesis to give higher priority to the equality operator. Check the priority precedence. By default, equality operators have least priority.
This question already has answers here:
What is the difference between prefix and postfix operators?
(13 answers)
Closed 7 years ago.
I don't understand how the results of this is :
2
2
2
Here is my code :
#include <stdio.h>
int main()
{
int a = 1, b = 1, x = 0, y = 0;
double w;
x = 1 + a++;
printf("x = %d\n", x);
printf("a = %d\n", a);
y = ++b;
printf("y = %d\n", y);
printf("b = %d\n", b);
}
Ok i understood the postfix and prefix but i still don't understand why a and b are 2 and not 1 .
They are not being saved anywhere
so when you say x=1+a++ and y=++b,
b becomes 2 and is saved in y . How does b keeps being 2 when not saved anywhere such as b=++b .
Sorry i am not sure if you guys follow what i am thinking.
You have to understand how the increment operator works.
You have two operations :
X++ => Return the value of X, then increment it by 1.
++X => Increment X by 1 then return it.
In your situation, the line with a problem is here : x = 1 + a++;
This translates into :
Return the value of a (1) and increment it (a becomes 2).
Set the value of x equal to the 1 + the value returned by a (1) (x becomes 2)
Hope this helps.
C Language Pre-increment and Post-increment Operators
#include <stdio.h>
int main()
{
int a = 1, b = 1, x = 0, y = 0;
double w;
x = 1 + a++;
printf("x = %d\n", x);
printf("a = %d\n", a);
y = ++b;
printf("y = %d\n", y);
printf("b = %d\n", b);
}
Pre-increment means increment variable before its value is used. Post-increment means increment variable after its value has been used. To understand how these operators work, let's look at the first case:
a = 1;
x = 1 + a++; // ++ on right means post-increment
First you're setting the value of a to 1. Then you're saying add a (value is 1) to x (value is 1). The result is x has a value of 2. And then, after the statement is done executing, as a side-effect, a is incremented, because you used the post-increment form of ++. So after x has been set to a value of 2, the value of a will become 2.
Instead, if you used a a pre-increment operator:
a = 1;
x = 1 + (++a); // ++ on left means pre-increment
Again, you start with a = 1, but this time, a is incremented before its value is used in the statement, because ++ on the left-side means pre-increment. In other words, first, a is incremented to the value of 2. Then a is added to x (whose value is 1), setting the value of x to 3.
In both the above cases, a starts out with a value of 1 and becomes 2. The difference is whether that happens before or after the value of a is used in the expression.
Can I compare three variables like the following, instead of doing if((x==y)&&(y==z)&&(z=x))? [The if statement should execute if all three variables have the same value. These are booleans.]
if(debounceATnow == debounceATlast == debounceATlastlast)
{
debounceANew = debounceATnow;
}
else
{
debounceANew = debounceAOld;
}
No, it does not.
x == y is converted to int, yields 0 or 1, and the result is compared to z. So x==y==z will yield true if and only if (x is equal to y and z is 1) or (x is not equal to y and z is 0)
What you want to do is
if(x == y && x == z)
No. The equality check associates from the left and the logical result is compared as a number, so that the expression 2 == 2 == 1 parses as (2 == 2) == 1, which in turn gives 1 == 1 and results in 1, which is probably not what you want.
You can actually type something like this:
int main()
{
const int first = 27,
second = first,
third = second,
fourth = third;
if (!((first & second & third) ^ fourth))
return 1;
return 0;
}