main()
{
unsigned int k = 987 , i = 0;
char trans[10];
do {
trans[i++] =(char) (k%16 > 9 ? k%16 - 10 + 'a' : '\0' );
} while(k /= 16);
printf("%s\n", trans);
}
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Answer Posted / lakshman
first case k=987, i=0.
trans[i++]=(char)(k%16>9?k%16-10+'a':''); means
trans [0]=char(987%16(=61)>9)condition is true so k%16-10+'a' value is print. that value is equal to 61-10|+(ascii value of a, that is equal to 97) so finally we get 98 and that is equal to b,
second case k=61, i=1 this will gives d
third case k=3, i=2 this gives null character.
the final answer is bd
| Is This Answer Correct ? | 1 Yes | 0 No |
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