Can't understand this if statement - c

Question

I've a C university exam coming up next week and i was looking at old exam papers a one of the questions gives this fragmented bit of code.
int a=2, b=-1, c=0;
if (a-2||b&&c||a){
printf("True\n");
} else {
printf("False\n");
}
We have to determine what the output of this code will be but the if statement makes no sense to me any if statement I've come across has been very specific like saying
if( x == 0)
I don't know what this is looking for my only assumption is that its going to be always true. Am I right or is there more to it then that?

Answer 1

This assignment has two goals:
to show what booleans are in C: Essentially they evaluate to ints with false mapping to 0 and true mapping to 1. In turn, any numeric or pointer value can be used in an integer context, with the respective zero value (0, 0.0, NULL (pointer), 0.0f, 0L etc.) evaluating as false and all others as true.
to show the precedence of operators
&& has a higher precedence than ||, so this statement is equivalent to
a-2 || (b&&c) || a
which will evaluate to true if any of the values is true.
As a==2, a-2 is 0. c is 0, so b && c is 0 as well.
So we have 0 || 0 || a, which is true as a is 2.

Answer 2

Most languages interprets non-zero integers as true and zero as false, so here you would have to calculate each one of the terms. Without any parenthesis, I would suggest that the && statement is taken in account first. So we have:
if (2-2 // gives zero
|| // OR
-1 && 0 // -1 AND 0 gives false
|| // OR
a) // Which is 2, which is true
So you're right, this statement is always true. This exercice was about showing predecence orders, and the fact that everything is numerical, even in boolean logic.
This is really important for you to understand.
If the predecence was the other way around (|| > &&), you must understand that it would have been false instead. I think this example's whole point is here.
(a-2 || b) && (c || a)
false && true
--> false

Answer 3

You need to understand that truth and falsity in C is always numerical.
https://www.le.ac.uk/users/rjm1/cotter/page_37.htm
Namely, anything that evaluates to numerical zero is false, and anything that evaluates to numerical non-zero is true.

Answer 4

In c language integers 0 is treated as false and any non-zero integer value is true but it should be noted that it is language specific and the sme statement will show compilation error in java as java is more strict and integers are not converted to booleans.
Talking about the above assignment problem the expression inside if statement will evaluate to true as
(a-2||b&&c||a) is same as
(2-2||-1&&0||2) which is same as
(0||0||2) which is evaluated as
(false||false||true) and hence the entire expression evaluates to
true.
hope it helps.

Answer 5

int a=2, b=-1, c=0;
int first=a-2; //0 -> false
bool second= b&& c; // nonZero&&zero -> true&&false -> false
int third = 2; // nonZero -> true
// false|| false|| true -> true
if (first || second || third ){
printf("True\n");
} else {
printf("False\n");
}

Answer 6

you need to understand two things before solving this problem that is
operator precedence and
associativity of operators
operator precedence tells c compiler that which operation to perform first.
and if two operators have same precedence than associativity tells evaluate left to right or right to left in you expression
int a=2, b=-1, c=0;
if (a-2||b&&c||a){
you can think it as
if((a-2)||(b&&c)||a){}
means - has top precedence so it will solved first
reduced to if(0||(b&&c)||a){}
then && has higher precedence so
reduced to if(0||false||a)
then the associativity is left to right so
reduced to if(false||a)
that is(false||2)
return true

Answer 7

In almost every programming language as far as I know 0 means false and 1 means true.
So coming up to your question: you have used && and || operators. Both of these are called Logical operators.
Now first block of yours is a-2||b :-
2-2||-1 so 0||-1. Now since the right expression of || is -1 the or operator will return 1 i.e. True because one of the values of 0 and -1 is non-zero 0 i.e. -1.
Therefore the expression resolves to 1&&c||a :-
Now c=0, therefore 1&&0 returns a 0 because && will only return 1 if both the expressions right and left of it are non zero.
So expression becomes 0||2 :-
Now since || (or operator) requires only one of operands either on right or left side to be non zero hence 0||2 returns 1.
Now your if (a-2||b&&c||a) statement resolves to
if (1)
{
printf("True\n"); }
else......
Therefore since 1 means TRUE the if statement will execute and you will get output as True.