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第1题: (b)volatile字面意思是易于挥发的。这个关键字来描述一个变量时，意味着 给该变量赋值(写入)之后，马上再读取，写入的值与读取的值可能不一样,所以说它"容易挥发"的。
这是因为这个变量可能一个寄存器，直接与外部设备相连，你写入之后，该寄存器也有可能被外部设备的写操作所改变;或者，该变量被一个中断程序，或另一个进程
改变了.
volatile variable isn't affected by the optimization. Its value after the longjump is the last value variable assumed.
b last value is 5 hence 5 is printed.
setjmp : Sets up for nonlocal goto /* setjmp.h*/
Stores context information such as register values so that the lomgjmp function can return control to the statement following the one calling setjmp.Returns 0 when it is initially called.
Lonjjmp: longjmp Performs nonlocal goto /* setjmp.h*/
Transfers control to the statement where the call to setjmp (which initialized buf) was made. Execution continues at this point as if longjmp cannot return the value 0.A nonvolatile automatic variable might be changed by a call to longjmp.When you use setjmp and longjmp, the only automatic variables guaranteed to remain valid are those declared volatile.
Note: Test program without volatile qualifier (result may very)
第2题: (a)
The members of structures have address in increasing order of their declaration. If a pointer to a structure is cast to the type of a pointer to its first member, the result refers to the first member.
第3题: (a)
Non recursive version of the program
int what ( int x , int n)/* Code raise a number (x) to a large power (n) using binary doubling strategy */
{
int val;
int product;
product =1;
val =x;
while(n>0)
{
if (n%2 == 1)
product = product*val;
n = n/2;
val = val* val;
}
}
Algorithm description
(while n>0)
{
if next most significant binary digit of n( power) is one
then multiply accumulated product by current val ,
reduce n(power) sequence by a factor of two using integer division .
get next val by multiply current value of itself
}
第4题: (c)
type of a is array of int
type of &a is pointer to array of int
Taking a pointer to the element one beyond the end of an array is sure to work.
第5题: (b)
第6题: (c)
The comma separates the elements of a function argument list. The comma is also used as an operator in comma expressions. Mixing the two uses of comma is legal, but you must use parentheses to distinguish them. the left operand E1 is evaluated as a void expression, then E2 is evaluated to give the result and type of the comma expression. By recursion, the expression
E1, E2, ..., En
results in the left-to-right evaluation of each Ei, with the value and type of En giving the result of the whole expression.
c=a,b; / *yields c=a* /
d=(a,b); /* d =b */
第7题: (a)
/* ptr is pointer to array of 3 int */
第8题: (c)
f1 and f2 return address of local variable ,when function exit local variable disappeared
第9题: (b)
sizeof operator gives the number of bytes required to store an object of the type of its operand . The operands is either an expression, which is not evaluated ( (++i + ++ i ) is not evaluated so i remain 3 and j is sizeof int that is 2) or a parenthesized type name.
第10题: (a)
void(*p[2]) ( int *, int);
define array of pointer to function accept two argument that is pointer to int and return int. p[0] = f1; p[1] = f2 contain address of function .function name without parenthesis represent address of function Value and address of variable is passed to function only argument that is effected is a (address is passed). Because of call by value f1, f2 can not effect b
第11题: (a)
第12题: (c)
C provide a facility called typedef for creating new data type names, for example declaration
typedef char stringMakes the name string a synonym for int .The type string can be used in declaration, cast, etc, exactly the same way that the type int can be. Notice that the type being declared in a typedef appears in the position of a variable name not after the word typedef.
第13题: (c)
If the type of an expression is "array of T" for some type T, then the value of the expression is a pointer to the first object in the array, and the type of the expression is altered to "pointer to T"
So (buf+1)[5] is equvalent to *(buf +6) or buf[6]
第14题: (b)
p+=sizeof(int) point to argv[2]
(p+=sizeof(int))[-1] points to argv[1]
第15题: (c)
When we call ripple value of the first argument passed to ripple is collected in the n that is 3. va_start initialize p to point to first unnamed argument that is 5 (first argument).Each call of va_arg return an argument and step p to the next argument. va_arg uses a type name to determine what type to return and how big a step to take Consider inner loop
(; i; i&=i-1) k++ /* count number of 1 bit in i *
in five number of 1 bits is (101) 2
in seven number of 1 bits is (111) 3
hence k return 5
example
let i= 9 = 1001The right most 1 bit of i has corresponding 0 bit in i-1 this way i & i-1, in a two complement number system will delete the right most 1 bit I(repeat until I become 0 gives number of 1 bits)
i-1 = 1000
(i-1) +1 = i
1000
+1
1 001
第16题: (b)
The answer is (b)
Static variable count remain in existence rather than coming and going each time function is called
so first call counter(0) count =0
second call counter(1) count = 0+1;
third call counter(2) count = 1+2; /* count = count +i */
fourth call counter(3) count = 3+3;
fifth call counter(4) count = 6+4;
sixth call counter(5) count = 10+5;
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