What is the output of this C code?
void main()
{
int k;
for (k = -3; k < -5; k++)
printf("Hello");
}
Answer: D
Correct answer: Nothing
A for tests its condition before the first pass. With k at -3, the test k < -5 is already false, so the body never runs and nothing is printed. (Had it been a do-while, the body would have run once.)
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What is the output of the following C code?
#include <stdio.h>
int main()
{
int index;
for(index=1; index<=5; index++)
{
printf("%d", index);
if (index==3)
continue;
}
}
Answer: B
Correct answer: 12345
The continue is the last statement in the body, so it skips nothing - the next thing after it would have been the end of the iteration anyway. Every pass prints its index, giving 12345.
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A char variable can store either an ASCII character or a Unicode character.
Answer: A
Correct answer: False
In C a char is one byte - sizeof(char) is 1 by definition - and on ordinary implementations that is eight bits, giving 256 possible values. That is ample for ASCII, which uses only the codes 0 to 127, but Unicode defines over a million code points, so a single char cannot hold an arbitrary Unicode character. C provides other types for that work: wchar_t for wide characters, and since C11 char16_t and char32_t for UTF-16 and UTF-32 code units, declared in <uchar.h>. What a char array can hold is a Unicode character encoded in UTF-8, but that takes one to four char elements per character, not one - the character is spread across several bytes rather than stored in a single variable. The confusion usually comes from Java, where char is a 16-bit UTF-16 code unit and the statement would be much closer to true. In C it is false.
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A short integer is at least 16 bits wide and a long integer is at least 32 bits wide.
Answer: A
Correct answer: True
The standard sets minimum widths, not exact ones: short must be at least 16 bits and long at least 32, with int at least 16 and long long at least 64. It also requires the ordering char ≤ short ≤ int ≤ long ≤ long long. Actual widths vary by platform, which is why <stdint.h> offers fixed-width types.
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#include <stdio.h>
int i;
int main()
{
if (i);
else
printf("Else");
return 0;
}
What is correct about the above program?
Answer: B
Correct answer: Else block is executed
i is a global, so it is zero-initialised - not indeterminate, which is what makes the outcome predictable. The condition is therefore false. The semicolon after if (i) is the empty statement that forms the if branch, and the else attaches to it perfectly legally, so the else branch runs.
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Output?
#include <stdio.h>
int main()
{
int c = 5, no = 10;
do {
no /= c;
} while(c--);
printf ("%d\n", no);
return 0;
}
Answer: B
Correct answer: Runtime Error
The loop divides no by c and then tests c--, so c falls 5, 4, 3, 2, 1, 0. On the pass after the test that saw c equal to 1, c is 0 - and no /= 0 is integer division by zero, which the hardware traps. The compiler cannot see it, so it is a run-time fault.
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What is the output of the below program?
#include <stdio.h>
int main()
{
int i = 0;
switch (i)
{
case '0': printf("Geeks");
break;
case '1': printf("Quiz");
break;
default: printf("GeeksQuiz");
}
return 0;
}
Answer: C
Correct answer: GeeksQuiz
'0' is a character constant, and its value is the ASCII code 48, not the number 0. Likewise '1' is 49. The controlling expression is 0, which matches neither, so control falls to default.
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When 2 is entered, The output of the code below is?
#include <stdio.h>
void main()
{
int ch;
printf("enter a value btw 1 to 2:");
scanf("%d", &ch);
switch (ch)
{
case 1:
printf("1\n");
break;
printf("Hi");
default:
printf("2\n");
}
}
Answer: D
Correct answer: 2
ch is 2, which matches no case, so control goes straight to default and prints 2. The printf("Hi") is unreachable in any event: it sits immediately after a break and before the next label, so no path can reach it.
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Which of the following cannot be used as LHS of the expression in for (exp1;exp2; exp3) ?
Answer: D
Correct answer: macros
The three parts of a for are ordinary expressions, so anything assignable can appear on the left of an assignment in them - a variable, or the result of a function returning a modifiable lvalue. A macro is only text, replaced before compilation, so what appears in the code is whatever it expands to; it is not itself an object and cannot be assigned to.
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Which of the following statement is correct for switch controlling expression?
Answer: B
Correct answer: Both int and char can be used in “switch” control expression.
The controlling expression must have an integer type. char qualifies - it is a small integer type and is promoted to int - as do short, long and enumerations. float and double do not, because equality against a case label would be unreliable for values that cannot be represented exactly. The expression also cannot be omitted.
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