OOP344 20093 Quiz 2
1 – What is the value of X?
int X = printf("A: %d B: %.2lf\n", 123, 123.456);
A: 2 B: -1 C: 1 D: 16 E: 17
2 – What is the value of X?
int a;
char b;
int X = scanf("%c%c%d%d%c", &b, &b, &a, &a, &b);
if user enters: A123BC
A: -1 B: 3 C: 4 D: 5 E: 6
3 – If
void* ptr = (void*)1234500;
ptr++;
Then value of ptr will be:
A: 1234501 B: 1234502 C: 1234504 D: 1234508 E: The code will not compile
4 – If
short int* ptr = (short int*)1234500;
ptr++;
Then value of ptr will be:
A: 1234501 B: 1234502 C: 1234504 D: 1234508 E: The code will not compile
5 – Having the following statement:
The number “123” as an address for a short integer.
We can surly say:
A: It is valid because 123 is an integer and like any other integer number, it could be and address for a short integer.
B: It is invalid because it too small to be an address
C: It is invalid because it is an odd number
D: It is invalid because it is not coefficient of 2, that is the size of a short integer.
E: It is invalid because of all said in C and D.
6 – Having
char ch = 127;
ch++;
The value of ch will be:
A: 128 B: -128 C: 0 D: -127 E: -1
Having:
int X[2][3][2] = { {{1,2},{3,4},{5,6} }, {{7,8},{9,10},{11,12} } };
int* p;
And knowing that X is located at address 12300 in memory:
7 – the pointer notation for X[0][2][1] is:
A: *(*(*(X+0)+2)+1)
B: *(*(*(X)+2)+1)
C: *(*(*X+2)+1)
D: *(*(*(X+1)+2)+0)
E: All A, B and C
8 - (X+1) is:
A: 12324
B: 12312
C: 12306
D: 12305
E: 12304
9 – (&X[0][0] + 1) is:
A: 12301
B: 12302
C: 12304
D: 12308
E: None of the above choices
10 – if p = (int*) X; and p++;
then “p” will be:
A: 12324
B: 12312
C: 12306
D: 12305
E: 12304