c语言高级面试题.doc

整个测试遵循以下的约定： 假定在所有的程序中必须的头文件都已经被正确包含。考虑如下的数据类型： char 为 1 个字节 int 为 4 个字节 long int 为 4 个字节 float 为 4 个字节 double 为个 8 字节 long double 为 8 个字节 指针为 4 个字节1. Consider the following program:#includestatic jmp_buf buf;main()volatile int b;b =3;if(setjmp(buf)!=0) printf(“%d “, b); exit(0);b=5;longjmp(buf , 1);The output for this program is: (a) 3 (b) 5 (c) 0 (d) None of the above2. Consider the following program:main()struct node int a;int b;int c; ;struct node s= 3, 5,6 ;struct node *pt = printf(“%d“ , *(int*)pt);The output for this program is: (a) 3 (b) 5 (c) 6 (d) 73. Consider the following code segment:int foo ( int x , int n)int val;val =1;if (n>0) if (n%2 = 1) val = val *x;val = val * foo(x*x , n/2);return val;What function of x and n is compute by this code segment? (a) xn (b) x*n (c) nx (d) None of the above4. Consider the following program:main()int a5 = ;int *ptr = (int*)(printf(“%d %d“ , *(a+1), *(ptr-1) );The output for this program is: (a) 2 2 (b) 2 1 (c) 2 5 (d) None of the above 5. Consider the following program:void foo(int 3 ); main()int a 33= 1,2,3 , 4,5,6,;foo(a);printf(“%d“ , a21);void foo( int b3)+ b;b11 =9;The output for this program is: (a) 8 (b) 9 (c) 7 (d) None of the above6. Consider the following program:main()int a, b,c, d;a=3;b=5;c=a,b;d=(a,b);printf(“c=%d“ ,c);printf(“d=%d“ ,d);The output for this program is: (a) c=3 d=3 (b) c=5 d=3 (c) c=3 d=5 (d) c=5 d=57. Consider the following program:main()int a3 = 1,2,3 ,4,5,6;int (*ptr)3 =a;printf(“%d %d “ ,(*ptr)1, (*ptr)2 );+ptr;printf(“%d %d“ ,(*ptr)1, (*ptr)2 );The output for this program is: (a) 2 3 5 6 (b) 2 3 4 5 (c) 4 5 0 0 (d) None of the above8. Consider following functionint *f1(void)int x =10;return(int *f2(void)int*ptr;*ptr =10;return ptr;int *f3(void)int *ptr;ptr=(int*) malloc(sizeof(int);return ptr;Which of the above three functions are likely to cause problem with pointers (a) Only f3 (b) Only f1 and f3 (c) Only f1 and f2 (d) f1 , f2 ,f39. Consider the following program:main()int i=3;int j;j = sizeof(+i+ +i);printf(“i=%d j=%d“, i ,j);The output for this program is:(a) i=4 j=2 (b) i=3 j=2 (c) i=3 j=4 (d) i=3 j=610. Consider the following program:void f1(int *, int); void f2(int *, int); void(*p2) ( int *, int);main()int a;int b;p0 = f1;p1 = f2;a=3;b=5;p0(printf(“%dt %dt“ , a ,b);p1(printf(“%dt %dt“ , a ,b);void f1( int* p , int q)int tmp;tmp =*p;*p = q;q= tmp;void f2( int* p , int q)int tmp;tmp =*p;*p = q;q= tmp; The output for this program is: (a) 5 5 5 5(b) 3 5 3 5 (c) 5 3 5 3 (d) 3 3 3 311. Consider the following program:void e(int ); main()int a;a=3;e(a);void e(int n)if(n>0)e(-n);printf(“%d“ , n);e(-n);The output for this program is: (a) 0 1 2 0(b) 0 1 2 1 (c) 1 2 0 1 (d) 0 2 1 112. Consider following declarationtypedef int (*test) ( float * , float*)test tmp;type of tmp is (a) Pointer to function of having two arguments that is pointer to float (b) int (c) Pointer to function having two argument that is pointer to float and return int (d) None of the above 13. Consider the following program:main()char *p;char buf10 = 1,2,3,4,5,6,9,8;p = (buf+1)5;printf(“%d“ , p);The output for this program is: (a) 5 (b) 6 (c) 9 (d) None of the above14. Consider the following program:Void f(char*);main()char * argv = “ab“ ,“cd“ , “ef“ ,“gh“, “ij“ ,“kl“ ;f( argv );void f( char *p )char* t;t= (p+= sizeof(int)-1;printf( “%s“ , t);The output for this program is: (a) ab (b) cd (c) ef (d) gh15. Consider the following program:#includeint ripple ( int , .);main()int num;num = ripple ( 3, 5,7);printf( “ %d“ , num);int ripple (int n, .)int i , j;int k; va_list p;k= 0;j = 1;va_start( p , n); for (; j0)if (n%2 = 1) product = product*val;n = n/2; val = val* val;/* Code raise a number (x) to a large power (n) using binary doubling strategy */ Algorithm description (while n>0) if next most significant binary digit of n( power) is onethen 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 Answer 4.The answer is (c)type of a is array of int type of / *yields c=a* /d=(a,b); /* d =b */Answer 7.The answer is (a)/* ptr is pointer to array of 3 int */ Answer 8.The answer is (c)f1 and f2 return address of local variable ,when function exit local variable disappeared Answer 9.The answer is (c)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. Answer 10.The answer is (a)void(*p2) ( int *, int); define array of pointer to function accept two argument that is pointer to int and return int. p0 = f1; p1 = 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 Answer 11.The answer is (a)Answer 12.The answer is (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. Answer 13.The answer is (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 buf6 Answer 14.The answer is (d)p+=sizeof(int) point to argv2 (p+=sizeof(int)-1 points to argv1 Answer 15.The answer is (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 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;