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https://gateoverflow.in/?qa=blob&qa_blobid=14433986388826671915

int main()
{
int a = 10;
int *b = &a;
scanf("%d",b);
printf("%d",a+50);
}

What will be the Output of the following code if input given is $25$ ?

edited

1 vote
int main() {
int a=10;            // value of a =10;
int *b=&a;          // pointer b stores the address of a i.e. points to a
scanf("%d",b);     // b gets a value from user(25) and stores it in a since b points to a
printf("%d",a+50);  // here a = (value which user gave + 50)= 25+50 = 75
return 0;
}

$\therefore$ here value of $a$ = (value which user gave + 50) = 25+50 = $75$ will be printed as output.

0
I'm having a doubt that b contains address of a than how can we update the value of a as=25 rather than not updating any value
Plz help with an approach for my doubt.
1

scanf reads data from stdin(here it is 25) and stores them according to the parameter format (here it is %d) into the locations pointed by the additional arguments.(i.e stores in location pointed by b)

we generally write scanf as scanf("%d" , &a); and here b = &a;

So we can write scanf("%d",b);

Output 75

The output will be 75 if input given is 25
Here the answer will be 75 because here you are putting the address of a in scanf. so, when scanf is called it will read the input value to the memory location a. so, in a instead of 10 25 will be stored and when 25 is added to 50 we will get answer as 75 which is printed as given in the code.

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