main() { char c; int i = 456; clrscr(); c = i; printf("%d", c); } - it takes more than one byte. Se when we assign c=i, the most significant will be lost. Then the value of c is 11001000. when you print the value of c using %d(signed integer), the most significat bit will be considered as the +/-. Here it is 1 and hence the number is negative. Negative numbers are in 2's compliment form. so converting back, we will get 00111000. ie, -56 in decimal. if you used %u instead of %d, you could have got 200.

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Question: main()
{
char c;
int i = 456;
clrscr();
c = i;
printf("%d", c);
}

Answer: it takes more than one byte. Se when we assign c=i, the most significant will be lost. Then the value of c is 11001000. when you print the value of c using %d(signed integer), the most significat bit will be considered as the +/-. Here it is 1 and hence the number is negative. Negative numbers are in 2's compliment form. so converting back, we will get 00111000. ie, -56 in decimal. if you used %u instead of %d, you could have got 200.

Category	C Interview Questions & Answers - Exam Mode / Learning Mode
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Question: main()
{
char c;
int i = 456;
clrscr();
c = i;
printf("%d", c);
}

Answer:

it takes more than one byte. Se when we assign c=i, the most significant will be lost. Then the value of c is 11001000. when you print the value of c using %d(signed integer), the most significat bit will be considered as the +/-. Here it is 1 and hence the number is negative. Negative numbers are in 2's compliment form. so converting back, we will get 00111000. ie, -56 in decimal. if you used %u instead of %d, you could have got 200.
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