# OOP344 Quiz2 20093

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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