1.下面这个表达式的类型和值分别是什么？
(float)(25 / 10);
#include
#include
int main( void ){
printf( "sizeof( (float)(25 / 10) ) = %zd, (float)(25 / 10) = %f, "
"sizeof( float ) = %zdn", sizeof( (float)(25 / 10) ), (float)(25 / 10), sizeof( float ) );
return EXIT_SUCCESS;
}
2.下面这个程序的结果是什么？
int func( void ){
static int counter = 1;
return ++counter;
}
int main( void ){
int answer;
answer = func() - func() * func();
printf( "%dn", answer );
return EXIT_SUCCESS;
}
解析：
这是一个狡猾的问题。比较明显的回答是-10(2 - 3 * 4)，但实际上它因编译器而异。乘法运算必须在加法运算之前完成，但并无规则规定函数调用完成的顺序。因此，下面几个答案都是正确的：
-10 (2 - 3 * 4) or (2 - 4 * 3)
-5 (3 - 2 * 4) or (3 - 4 * 2)
-2 (4 - 2 * 3) or (4 - 3 * 2)
下面是DEV-C++编译器的答案：
#include
#include
int func( void ){
static int counter = 1;
return ++counter;
}
int main( void ){
int answer;
answer = func() - func() * func();
printf( "%dn", answer );
return EXIT_SUCCESS;
}
3.位操作符和移位操作符可以用在什么地方？
解析：
位操作符可以用在让某一位置为1，或者让某一位清零。
移位操作符可以用在进行更快的乘除法，两者结合起来可以用来查看值在内存中的表现形式。
4.条件操作符在运行时比较时较之if语句是更快还是更慢？试着比较下面两个代码段。
if( a > 3 ){
i = b + 1;
} else{
i = c * 5;
}
i = a > 3 ? b + 1 : c * 5;
解析：
不，它们都执行相同的任务。如果你比较吹毛求疵，使用if的哪个方案看上去稍微臃肿一些，因为它具有两条存储到i的指令。但是，它们之间只有一条指令才会执行，所以在速度上并无差别。
#include
#include
int main( void ){
int a = 5;
int b = 6;
int c = 2;
int i;
if( a > 3 ){
i = b + 1;
} else{
i = c * 5;
}
printf( "1:use if statement:i = %d, a = %d, b = %d, c = %dn", i, a, b, c );
i = a > 3 ? b + 1 : c * 5;
printf( "2:use ?: operator:i = %d, a = %d, b = %d, c = %dn", i, a, b, c );
return EXIT_SUCCESS;
}
5.可以被4整除的年份是闰年，但是其中能够被100整除的年份又不是闰年。但是，这其中能够被400整除的年份又是闰年。请写一条赋值语句，实现如下功能：如果变量year的值是闰年，把变量leap_year设置为真；如果year的值不是闰年，把leap_year设置为假。
#include
#include
int main( void ){
int leap_year = 0;
int year;
year = 2900;
if( year % 4 == 0 ){
leap_year = 1;
}
if( year % 4 == 0 && year % 100 == 0 ){
leap_year = 0;
}
if( year % 4 == 0 && year % 100 != 0 ){
leap_year = 1;
}
if( year % 4 == 0 && year % 100 == 0 && year % 400 == 0 ){
leap_year = 1;
}
printf( "method01: %d year is %s year.n", year, leap_year == 1 ? "leap" : " not leap " );
if( year % 400 == 0 ){
leap_year = 1;
} else if( year % 100 == 0 ){
leap_year = 0;
} else if( year % 4 == 0 ){
leap_year = 1;
} else{
leap_year = 0;
}
printf( "method02: %d year is %s year.n", year, leap_year == 1 ? "leap" : " not leap " );
return EXIT_SUCCESS;
}
6.哪些操作符具有副作用？它们具有什么副作用？
解析：
()操作符并无任何副作用，但它所调用的函数可能有副作用。
操作符 副作用
++,-- 不论是前缀还是后缀形式，这些操作符都会修改它们的操作数
= 包括所有其它的复合赋值符：它们都修改作为左值的左操作数
#include
#include
int main( void ){
int a = 1, b = 1;
int prefix_a;
int b_post;
printf( "a = %d, b = %dn", a, b );
prefix_a = ++a;
printf( "after prefix_a = ++a, a = %d, prefix_a = %dn", a, prefix_a );
b_post = b++;
printf( "after b_post = b++, b = %d, b_post = %dn", b, b_post );
return EXIT_SUCCESS;
}
7.下面这个代码段的结果是什么？
int a = 20;
...
if( 1 <= a <= 10 )
printf( "In rangen" );
else
printf( "Out of rangen" );
#include
#include
int main( void ){
int a = 20;
if( 1 <= a <= 10 )
printf( "In rangen" );
else
printf( "Out of rangen" );
return EXIT_SUCCESS;
}
8.改写下面的代码段，消除多余的代码。
a = f1( x );
b = f2( x + a );
for( c = f3( a, b ); c > 0; c = f3( a, b ) ){
statements
a = f1( ++x );
b = f2( x + a );
}
解析：
#include
#include
int main( void ){
for( c = f3( f1( x ), f2( x + a ) ); c > 0; c = f3( f1( ++x ), f2( x + a ) ) ){
statements;
}
return EXIT_SUCCESS;
}
9.下面的循环能够实现它的目的吗？
解析：
不能实现。因为并不是只有所有的零相加才为零。
#include
#include
#define ARRAY_SIZE 4
int main( void ){
int non_zero = 0;
int i;
int array[ARRAY_SIZE] = { 1, 2, -3, 0 };
printf( "print elements of array:n" );
for( i = 0; i < ARRAY_SIZE; i += 1 ){
printf("%d ", array[i] );
}
printf( "n" );
for( i = 0; i < ARRAY_SIZE; i += 1 ){
non_zero += array[i];
}
if( !non_zero ){
printf( "Values are all zeron" );
} else{
printf( "There are nonzero valuesn" );
}
return EXIT_SUCCESS;
}
10.根据下面的变量声明和初始化，计算下列每个表达式的值。如果某个表达式具有副作用（也就是说它修改了一个或多个变量的值）， 注明他们。在计算每个表达式时，每个变量所使用的是开始时给出的初始值，而不是前一个表达式的结果。
#include
#include
int main( void ){
int a = 10, b = -25;
int c = 0, d = 3;
int e = 20;
printf( "a = %d, b = %d, c = %d, d = %d, e = %dn", a, b, c, d, e );
printf( "a.b = %dn", b );
printf( "b.b++ = %dn", b++ );
printf( "after b++, b = %dn", b );
b = -25;
printf( "after b = -25, b = %dn", b );
printf( "c.--a = %dn", --a );
printf( "after --a, a = %dn", a );
a = 10;
printf( "after a = 10, a = %dn", a );
printf( "d.a / 6 = %dn", a / 6 );
printf( "e.a %% 6 = %dn", a % 6 );
printf( "f.b %% 10 = %dn", b % 10 );
printf( "g.a << 2 = %dn", a << 2 );
printf( "h.b >> 3 = %dn", b >> 3 );
printf( "i.a > b = %dn", a > b );
printf( "j.b = a = %dn", b = a );
printf( "after b = a, b = %dn", b );
b = -25;
printf( "after b = -25, b = %dn", b );
printf( "k.b == a = %dn", b == a );
printf( "l.a & b = %dn", a & b );
printf( "m.a ^ b = %dn", a ^ b );
printf( "n.a | b = %dn", a | b );
printf( "o.~b = %dn", ~b );
printf( "p.c && a = %dn", c && a );
printf( "q.c || a = %dn", c || a );
printf( "r.b ? a : c = %dn", b ? a : c );
printf( "s.a += 2 = %dn", a += 2 );
printf( "after a += 2, a = %dn", a );
a = 10;
printf( "after a = 10, a = %dn", a );
printf( "t.b &= 20 = %dn", b &= 20 );
printf( "after b &= 20, b = %dn", b );
b = -25;
printf( "after b = -25, b = %dn", b );
printf( "u.b >>= 3 = %dn", b >>= 3 );
printf( "after b >>= 3, b = %dn", b );
b = -25;
printf( "after b = -25, b = %dn", b );
printf( "v.a %%= 6 = %dn", a %= 6 );
printf( "after a %% 6, a = %dn", a );
a = 10;
printf( "after a = 10, a = %dn", a );
printf( "w.d = a > b = %d, a > b = %dn", d = a > b, a > b );
printf( "after d = a > b, d = %dn", d );
d = 3;
printf( "after d = 3, d = %dn", d );
printf( "x.a = b = c = dn", a = b = c = d );
printf( "after a = b = c = d, a = %d, b = %d, c = %dn", a, b, c );
a = 10;
b = -25;
c = 0;
printf( "after a = 10, b = -25, c = 0, a = %d, b = %d, c = %dn", a, b, c );
printf( "y.e = d + (c = a + b) + c = %dn", e = d + (c = a + b) + c );
printf( "after e = d + (c = a + b) + c, c = %d, e = %dn", c, e );
c = 0;
e = 20;
printf( "after c = 0, e = 20, c = %d, e = %dn", c, e );
printf( "z.a + b * 3 = %dn", a + b * 3 );
printf( "aa. b >> a - 4 = %dn", b >> a - 4 );
printf( "b >> (a - 4) = %dn", b >> (a - 4) );
printf( "bb.a != b != c = %d, a != b = %d, c = %dn", a != b != c, a != b, c );
printf( "cc.a == b == c = %d, a == b = %d, c = %dn", a == b == c, a == b, c );
printf( "dd.d < a < e = %d, d < a = %d, e = %dn", d < a < e, d < a, e );
printf( "ee.e > a > d = %d, e > a = %d, d = %dn", e > a > d, e > a, d );
printf( "ff.a - 10 > b + 10 = %d, a - 10 = %d, b + 10 = %dn", a - 10 > b + 10, a - 10, b + 10 );
printf( "gg.a & 0x1 == b & 0x1 = %d, a = %d, 0x1 == b = %d, a & 0x1 == b = %dn",
a & 0x1 == b & 0x1, a, 0x1 == b, a & 0x1 == b );
printf( "a & (0x1 == b) & 0x1 = %d, 0x1 == b = %d, a & (0x1 == b) = %dn",
a & (0x1 == b) & 0x1, 0x1 == b, a & (0x1 == b) );
printf( "hh.a | b << a & b = %d, a = %d, b << a = %d, b = %d, b << a & b = %dn", a | b << a & b, a, b << a,
b, b << a & b );
printf( "a | ((b << a) & b) = %dn", a | ((b << a) & b) );
printf( "ii.a > c || ++a > b = %dn", a > c || ++a > b );
printf( "after a > c || ++a > b, a > c = %d, a = %dn", a > c, a );
printf( "jj.a > c && ++a > b = %dn", a > c && ++a > b );
printf( "after a > c && ++a > b, a = %d, a > b = %dn", a, a > b );
a = 10;
printf( "after a = 10, a = %dn", a );
printf( "kk.!~b++ = %dn", !~b++ );
printf( "after !~b++, b = %dn", b );
b = --b;
printf( "after b = --b, b = %dn", b );
printf( "b = %d, ~b = %d, !~b = %dn", b, ~b, !~b );
b = -25;
printf( "after b = -25; b = %dn", b );
printf( "ll.b++ & a <= 30 = %d, a = %d, a <= 30 = %dn", b++ & a <= 30, a, a <= 30 );
printf( "after b++ & a <= 30, b = %dn", b );
b = --b;
printf( "after b = --b, b = %dn", b );
printf( "b & a <= 30 = %d, a = %d, a <= 30 =%dn", b && a <= 30, a, a <= 30 );
b = -25;
printf( "after b = -25, b = %dn", b );
printf( "mm.a - b, c += d, e - c = %d, a = %d, b = %d, d = %d, e = %dn", (a - b, c += d, e - c), a, b, d, e );
printf( "after a - b, c += d, e - c, c = %dn", c );
c = 0;
printf( "after c = 0; c = %dn", c );
printf( "nn.a <<= 3 > 0 = %d, 3 > 0 = %dn", a <<= 3 > 0, 3 > 0 );
printf( "after a <<= 3 > 0, a = %dn", a );
a = 10;
printf( "after a = 10, a = %dn", a );
printf( "a << (3 > 0) = %dn", a << (3 > 0) );
printf( "oo.a <<= d > 20 ? b && c++ : d-- = %dn", a <<= d > 20 ? b && c++ : d-- );
printf( "after a <<= d > 20 ? b && c++ : d--, d = %d, d > 20 = %d, a = %dn", d, d > 20, a );
return EXIT_SUCCESS;
}
11.下面列出了几个表达式。请判断编译器是如何对各个表达式进行求值的，并在不改变求值顺序的情况下，尽可能去除多余的括号。
#include
#include
int main( void ){
int a = 1, b = 6, c = 2;
printf( "a.a + (b / c) = %dn", a + (b / c) );
printf( "a + b / c = %dnn", a + b / c );
printf( "b.(a + b) / c = %dnn", (a + b) / c );
printf( "c.(a * b) %% 6 = %dn", (a * b) % 6);
printf( "c.a * b %% 6 = %dnn", a * b % 6);
printf( "d.a * (b %% 6) = %dnn", a * (b % 6) );
printf( "e.(a + b) == 6 = %dn", (a + b) == 6 );
printf( "a + b == 6 = %dnn", a + b == 6 );
printf( "f.!((a >= '0') && (a <= '9')) = %dn", !((a >= '0') && (a <= '9')) );
printf( "!(a >= '0' && a <= '9') = %dnn", !(a >= '0' && a <= '9') );
printf( "g.((a & 0x2f) == (b | 1)) && ((~c) > 0) = %dn", ((a & 0x2f) == (b | 1)) && ((~c) > 0) );
printf( "(a & 0x2f) == (b | 1) && ~c > 0 = %dnn", (a & 0x2f) == (b | 1) && ~c > 0 );
printf( "h.((a << b) - 3) < (b << (a + 3)) = %dn", ((a << b) - 3) < (b << (a + 3)) );
printf( "(a << b) - 3 < b << a + 3 = %dnn", (a << b) - 3 < b << a + 3 );
printf( "i.~(a++) = %dn", ~(a++) );
a = 1;
printf( "~a++ = %dnn", ~a++ );
printf( "j.((a==2) || (a == 4)) && ((b == 2) || (b == 4)) = %dn", ((a==2) || (a == 4)) && ((b == 2) || (b == 4)) );
printf( "(a==2 || a == 4) && (b == 2 || b == 4) = %dnn", (a==2 || a == 4) && (b == 2 || b == 4) );
printf( "k.(a & b) ^ (a | b) = %dn", (a & b) ^ (a | b) );
printf( "a & b ^ (a | b) = %dnn", a & b ^ (a | b) );
printf( "l.(a + (b + c)) = %dn", (a + (b + c)) );
printf( "a + b + c = %dnn", a + b + c );
return EXIT_SUCCESS;
}
12.在你的机器上对一个有符号值进行右移位操作时，如何判断执行的是算术移位还是逻辑移位？
解析：
通过查看内存中位的分布来判断进行算术移位还是逻辑移位。
#include
#include
#include
int main( void ){
int si;
int si2;
int bit_of_int;
int i;
si = -5;
bit_of_int = sizeof(int) * CHAR_BIT;
printf( "si = %dn", si );
printf( "print the bit of number %d in binary system:n", si );
for( i = bit_of_int - 1; i >= 0; --i ){
printf( "%c", si & (1u << i) ? '1' : '0' );
}
printf( "n" );
si2 = si >> 1;
printf( "after si2 = si >> 1, si2 = %d, print the bit of number %d in binary system:n", si2, si2 );
for( i = bit_of_int - 1; i >= 0; --i ){
printf( "%c", si2 & (1u << i) ? '1' : '0' );
}
printf( "n" );
return EXIT_SUCCESS;
}
