160 views

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 | 160 views

+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.

by Boss (23.7k points)
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 by (343 points)