笔试二三题(C语言)

题一： char c = 100; size_t i = sizeof(c = (c+155)); printf("%u %d/n", i, c); i = sizeof(int [c = (c + 155)]); printf("%u %d/n", i, c); i = sizeof(int [c = (c - 1)]); printf("%u %d/n", i, c);

这道题目的考察点是 sizeof 操作符。

在ANSI C ISO/IEC 9899:1999(E)里面，6.5.3.4节对 the sizeof operator做了如下说明：

The sizeof operator shall not be applied to an expression that has function type or an incomplete type, to the parenthesized name of such a type, or to an expression that designates a bit-field member.

The operand can be an expression or the parenthesized name of a type.

If the type of operand is a variable length array type, the operand is evaluated; otherwise, the operand is not evaluated and the result is an integer constant.

The value of result is implementation-define, and its type is size_t (an unsigned integer type), defined in <stddef.h>.

注：variable length array type为C99引入的新特性

按照这个说明，显然在Line 2，i的值在编译期使用常量替换了；而Line 4，因为sizeof的操作数是一个变长数组，故i值是在程序运行期确定的。

 

答案：

1  100

4294967292 -1

4294967288 -2

注：unsigned(-1 * 4) == 4294967292, unsigend(-2 * 4) == 4294967288

下面新增几个，作为思考题： int fun() { return 0; } void fun2() { printf("fun2 called/n"); } printf("%u %u %u %u/n", sizeof(fun), sizeof(&fun), sizeof(fun()), sizeof(fun2()));

 

 

 

题二：int a[5] = {1,2,3,4,5};int *ptr = (int*)(&a + 1);printf("%d, %d/n", *(a + 1), *(ptr - 1)); 该题考察点指针运算：

(1)、对一个变量取地址，则其类型为一个指向该变量类型的指针，这里&a，a为一个具有5个元素的整形数组，故&a获得的是一个指向具有5个元素的整形数组的指针。注：如果int a[5];int (*ptr0)[] = &a;int *ptr1 = &a;
在gcc编译环境下，Line 3会产生一个"warning: initialization from incompatible pointer type" ，原因很简单，ptr是一个整形指针，而&a不是

(2)、指针与常量的加减，这里三条语句是等效的：int *ptr = (int*)(&a + 1);int *ptr = (int*)((char*)&a + 1*sizeof(a)); int *ptr = (int*)&a + 1*sizeof(a)/sizeof(int);

注：两个指针变量相减，则必须要求二者为同一类型的指针，且无论两个指针变量是否为相同类型，均不能相加，否则编译无法通过，产生如下错误"invalid operands to binary -"

 

 

题三：int* func(){ static i = 0; i++; return &i;}void test1(){ int* a = func(); printf("%d %d/n", *a, *func());}void test2(){ printf("%d %d/n", *func(), *func());}int main(int argc, char* argv[]){ test1(); test2(); return 0;} 该题涉及的知识点有静态变量、函数参数压栈顺序(一般为从右向左压栈)、函数参数计算顺序(编译器相关)等。

注：stdcall、cdecl所表示的是函数调用约定，二者的参数压栈顺序均为从右向左；不同之处是cdecl调用约定，函数本身不清理堆栈，调用者负责清理堆栈，而stdcall则是被调用者自己恢复堆栈，故stdcall不能支持变长参数函数，因此printf这类函数只能是cdecl调用约定。

gcc 4.0.2编译环境下的答案为：

2 2
4 3

使用gcc编译出的汇编代码如下： .file "static.c" .local i.1655 .comm i.1655,4,4 .text.globl func .type func, @functionfunc: pushl %ebp movl %esp, %ebp movl i.1655, %eax incl %eax movl %eax, i.1655 movl $i.1655, %eax popl %ebp ret .size func, .-func .section .rodata.LC0: .string "%d %d/n" .text.globl test1 .type test1, @functiontest1: pushl %ebp movl %esp, %ebp subl $24, %esp call func movl %eax, -4(%ebp) call func movl (%eax), %edx movl -4(%ebp), %eax movl (%eax), %eax subl $4, %esp pushl %edx pushl %eax pushl $.LC0 call printf addl $16, %esp leave ret .size test1, .-test1.globl test2 .type test2, @functiontest2: pushl %ebp movl %esp, %ebp pushl %ebx subl $4, %esp call func movl (%eax), %ebx call func movl (%eax), %eax subl $4, %esp pushl %ebx pushl %eax pushl $.LC0 call printf addl $16, %esp movl -4(%ebp), %ebx leave ret .size test2, .-test2.globl main .type main, @functionmain: pushl %ebp movl %esp, %ebp subl $8, %esp andl $-16, %esp movl $0, %eax addl $15, %eax addl $15, %eax shrl $4, %eax sall $4, %eax subl %eax, %esp call test1 call test2 movl $0, %eax leave ret .size main, .-main .ident "GCC: (GNU) 4.0.2 20050901 (prerelease) (SUSE Linux)" .section .note.GNU-stack,"",@progbits   

 

题四：求结构体中某个成员变量的偏移量

1、gcc 4编译器中带有offsetof宏，利用该宏可直接求得偏移量，实例代码如下：#include <stddef.h>#include <stdio.h>int main(){ printf("%d/n", (int)offsetof(struct {int8_t TC_top; TC_a TC_bot;}, TC_bot)); return 0;}

2、自行实现offsetof宏，定义如下：#define offsetof(TYPE, MEMBER) ((size_t) & ((TYPE *)0)->MEMBER )