C for Everyone: Programming Fundamentals Final Exam (Week 6 Coursera)
In a format string for printf which would you use to print an int?
%d
In ANSI standard C(1989) code - this is what we are teaching- declarations can occur in a for statement as in for(int i; i < 10; i++0 ...
F
Assume the given declarations and fill in the value of the expression . int a = 1, b = 2, c = 3; Expression: a - b * c
-5
Assume the given declarations and fill in the value of the expression . int a = 3, b = 4, c = 0, d = '1'; Expression: a / b > c
0
Assume the given declarations and fill in the value of the expression . int a = 1, b = 2, c = 3; Expression: a++ + --b
2
Assume the given declarations and fill in the value of the expression . int a = 1, b = 2, c = 3; Expression: c / a * b
6
In the following code fragment int a[10] = {1,2,3,4,5,6,7,8,9,10}, i = 6 ; int *p = &a[0]; printf("%d\n", *(p + i)); what gets printed?
7
The expression PI * radius * radius would be used to compute
A circles area
if p and q are pointers to double and x and y are double which of the following is legal:
y = *q;
The input function scanf() uses "call by reference" to pass in argument values.
True
If a function's declaration is int foo(void):
The function has no arguments
With the code int foobar(int* n){ *n = *n +1; return *n; } when called int k = 6; printf("foobar(k) = %d,",foobar(&k) ); printf(" k = %d\n", k); what gets printed?
foobar(k) = 7, k = 7
Assume the given declarations and fill in the value of the expression . int a = 3, b = 4, c = 0, d = '1'; Expression: c < b && a > 3
0
Assume the given declarations and fill in the value of the expression . int i = 0, j = 1, k = 2; Expression: !!i
0
Assume the given declarations and fill in the value of the expression . int i = 0, j = 1, k = 2; Expression: i && j
0
Assume the given declarations and fill in the value of the expression . int i = 0, j = 1, k = 2; Expression: i || !k
0
Assume the given declarations and fill in the value of the expression . int a = 3, b = 4, c = 0, d = '1'; Expression: a < b
1
Assume the given declarations and fill in the value of the expression . int a = 3, b = 4, c = 0, d = '1'; Expression: b % a
1
Assume the given declarations and fill in the value of the expression. int i = 0, j = 1, k = 2; Expression: (i && (j = k)) || (k > j)
1
Assume the given declarations and fill in the value of the expression . int a = 1, b = 2, c = 3; Expression: b = a = c
3
Assume the given declarations and fill in the value of the expression . int a = 3, b = 4, c = 0, d = '1'; Expression: a % b
3
Assume the given declarations and fill in the value of the expression . int a = 3, b = 4, c = 0, d = '1'; Expression: c = a++
3
If you declare the array char mystr[10];
None of these
True/ False An if-else statement can always replace an if statement.
True
True/False A while can always be used to replace a for.
True
True/False: The declaration char* str = & a[0]; where char a[5] = "abcd"; The value of *str is the char 'a';
True
When you compile a correct C program you get a machine executable file such as a.out produced by the gnu compiler gcc.
True
The function mystery is defined as int mystery(int p, int q){ int r; if ((r = p % q) == 0) return q; else return mystery(q, r); } When called with mystery(7, 91) it will return
7
Who invented the C language?
Dennis Ritchie invented C at Bell Labs
The function mystery is defined as int mystery(int p, int q){ int r; if ((r = p % q) == 0) return q; else return mystery(q, r); } When called with mystery(2, 6) it will return
2
A variable declared in a inner block that has the same name as one in the surrounding block causes an error.
F
The following program is suppose to write Hello World onto the screen but it has syntax errors - find and correct. #include <stdio.h> int main(void) { printf(" Hello World\n"); return 0, }
It should be return 0;
The original intent of using register int i; was
It was intended to compile optimized code
The statement printf("HELLO\t\tWORLD\n");
Prints HELLO WORLD followed by a new line
What happens when the return statement has a double expression and the function return type is int?
There is a conversion from double to int
When declaring fact() why not use double fact() so that you do not have integer overflow?
There would be problems with accuracy.
Which is true: 1. #define is a preprocessor command often used to introduce named constants 2. double and goto are keywords declaring types. 3. return (0); is normally the last statement in main() 4. The file stdio.h is where the compiler finds scanf().
1, 3, 4
In the following code fragment what gets printed? int data[5] = {0 ,1, 2, 3, 4}, sum = 0 , i; for (i = 0; i < 5 ; i++) sum = sum + data[i]; printf("%d\n", sum);
10
The code i = -10; while ( i < 0) { ... do something ; i--; } Displays a common error
its an infinite loop
True/False - Factorial can only be computed recursively.
F
True/False mergesort is less efficient than bubblesort for very large sorting problems.
F
