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The C aptitude questions and answers for those questions along with explanation for the interview related queries.
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Read more about how to prepare and write the C Aptitude Exams?
Read more about C Programming Language . and read the C Programming Language (2nd Edition). by K and R.
Predict the output or error(s) for the following:
C aptitude 24.1
main()
{
char str[4]="test";
printf("%s",str);
}
Answer: test#$634278621@!
Explanation: The character array has the memory just enough to hold the string “test” and doesn’t have enough space to store the terminating null character. So it prints the test correctly and continues to print garbage values till it accidentally comes across a NULL character.
C aptitude 24.2
main()
{
char * str = "hello";
char * ptr = str;
char least = 127;
while (*ptr++)
least = (*ptr <least ) ?*ptr :least;
printf("%d",least);
}
Answer: 0
Explanation: After ‘ptr’ reaches the end of the string the value pointed by ‘str’ is ‘ ’. So the value of ‘str’ is less than that of ‘least’. So the value of ‘least’ finally is 0.
C aptitude 24.3
main()
{
int i = 257;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer: 11
Explanation: The integer value 257 is stored in the memory as, 00000001 00000001, so the individual bytes are taken by casting it to char * and get printed.
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