来看用指针实现高低位倒序！

gaosmile · 2020-05-13T11:29:19+00:00

昨晚在微信群看到一个读者发的面试题目，从网上截图出来的，我百思不得其解，题目如图。# w: W5 ?* h6 Z, v2 B & O8 }0 t: [' F" v幸好，我学过栈2 {% R! ]4 x. `) M 6 A; ~* y# X, R; S- T) G然后我写了个小程序/ Y+ z h1 f: c2 B 第一个方法比较笨，当我写完自己的代码后，看到有同学发了自己的代码，我赶紧就发了个红包，一个是为了鼓励大家多讨论问题，一个是为了赞扬这样的行为。 4 x H9 a+ \2 V+ i5 v z" b6 @第一个方法就是取出每一个bit，压栈，然后再出栈，你想怎么排序都没有问题。6 e$ e1 N# P5 }% l #include "stdio.h"* d0 r* h# b! W #include "stdlib.h" : k! U+ s9 v# {& x# D4 J. i0 y0 S. A8 w5 q+ | struct List{1 A; M, ]! B7 L7 M$ A K int data;! o# E7 A# ]1 B/ C2 `& y struct List * next; 9 K+ l: Q; V' b* h, N* l' `};# X, u# Y- P0 T0 L $ B& U& Y$ [0 L% M* z struct Stack{7 D7 z$ v4 ? m, M1 u: R struct List *head;4 g, U8 L8 k& N5 c% } int size;# U4 h5 B0 E; I6 K8 I1 L }; ' n( b K9 A( z: M6 V8 |8 H7 R" X 4 Q" v" w4 |6 q+ L struct Stack * StackInit(void)' F8 |# ]3 E* V { % Q* i- X9 n, M% A: Z' v! \ struct Stack *stack = NULL; ( J: V8 x( ?+ F8 H7 f stack = (struct Stack*)malloc(sizeof(struct Stack)); 4 e& P4 _" M; l/ H stack->head = (struct List *)malloc(sizeof(struct List));( H- I9 t% j" P! A H. c0 i0 u stack->head->next = NULL;+ P* T: a; k* v' P8 O1 |- _ stack->size = 0; 4 U% r3 z4 l/ ]; e$ s return stack; ' B6 D2 l5 c' G4 J j} ( i9 S% M. n' Q8 h3 q0 F& Z4 J3 s" k# s& c int StackPush(struct Stack *stack,int data) : Z3 m$ Y) J0 M# p- a* F{" u) b4 u( c! G$ F struct List *tmp = (struct List *)malloc(sizeof(struct List)); + w. p3 \1 ?& R8 ]9 ?8 z) { tmp->data = data; * j: X$ J( [ {1 s; j6 g8 s tmp->next = stack->head->next;! ?5 r* g% I# @+ L0 } stack->head->next = tmp; ; c! r- u) L3 p+ y( \ stack->size++;% l2 u' A; x: ^9 s1 b" G9 R% ] //printf("push:%d \n",data);4 i! @3 U# u& l& s return 0; ! L- r$ h: s, P' |( P} 5 ?4 |& B; M/ g" y0 b0 v, E# ~7 N7 T/ ^ int IsStackEmpty(struct Stack *stack) ' {+ g- m2 b$ |* `, v7 X{- k9 P; G8 n8 x! _+ J# J7 M if(stack->head->next == NULL){ 4 K0 j% d- L k w' O3 J //printf("stack null\n");3 U4 [. W* s+ F7 U return 1;. J: E+ z! ~- r B: {8 ^. z# f }else{2 Z8 T0 z1 _+ o6 [6 t- | //printf("stack is not null\n");8 B/ w/ ^9 W, E" Y" ~0 J8 w return 0;5 B# s3 V8 ~! q2 p( T' o" Y } 2 }+ B2 O% n- b* b} & z+ ?4 v* e; |: O7 {7 F , |" x: H: K. Zint StackPop(struct Stack *stack,int *data) - O: }. W% |, H6 X) f" ?{ * |6 P. e% O) Z5 s struct List *tmp = NULL; ~! {( C* u2 d# _ if(IsStackEmpty(stack)) 1 k c6 e2 ?0 m* v9 @; x+ w return -1; # _# \- |3 X6 Q) q6 G ; p% t& I2 |' i' l) J tmp = stack->head->next;8 I, _5 F1 B- E2 l7 a *data = tmp->data; : x- Y' @8 N) T: `0 @+ A+ z stack->head->next = tmp->next; 9 o$ s( J6 q5 h$ C0 s f1 w stack->size--; 0 E7 d K! j; ?' I/ q& P9 \2 y if(tmp!=NULL) & [6 Q7 x& \- y free(tmp); ) F# e1 m9 A$ ~% v/ y6 T 8 e! x9 J2 x- B* {, \ //printf("pop:%d \n",*data); " M- s: [5 k C: L: T* I$ h. R return 0;/ G( X6 g- v$ M }1 L8 z; m; v7 c/ l8 U& q h : n! y6 M) g. \# ]* y! Q( f " Q3 l) r8 \1 j; P int main(void) * {- U4 d1 S0 ]% _5 x# q{5 e' _/ i; }0 @: B/ f& { int i = 0;. x7 D! T5 {) K& H- b1 ~/ W' `$ t unsigned char a = 0x71;$ A5 ?- D: `3 Z3 W2 K+ P8 } unsigned char b = 0;- I9 u+ b" l' W, z+ ?5 T7 V " a$ t$ r6 |( b$ w6 _3 ?1 {. I struct Stack *stack = NULL; ) T4 b$ X* |6 g! I4 E9 w stack = StackInit();) ` R& W1 p! O# Z V0 z+ p. `# u2 Q9 b7 L" q printf("a:0x%x\n",a);2 T$ z6 z4 s8 p f- J/ u: v for(i = 0;i<8;i++): v$ L4 K4 K! Y+ j# b5 A4 x3 }, I { * D, D$ V$ w! @8 K; p% `$ v6 C( C " d# z: _) j8 L2 J2 O t if(a&0x10) StackPush(stack,1);/ l8 Z w' K! @1 k; H/ I0 Z9 \1 G) ? else StackPush(stack,0); : e3 k3 d ]. D6 k; Y2 Q3 m* ^* s if(a&0x20) StackPush(stack,1); / @0 M% J, T# H- r- I& w, k else StackPush(stack,0);! E/ v5 j& Y/ n$ a if(a&0x40) StackPush(stack,1);. n* w' E. g9 r) z( k else StackPush(stack,0); k4 S% N1 N3 O8 `" |+ c/ t if(a&0x80) StackPush(stack,1);8 i, p1 ~: A. z& Z$ U: `- r1 R3 L else StackPush(stack,0);. {; J, o5 L7 \# C0 S' {- } * |7 N: a, I% b: N if(a&0x01) StackPush(stack,1); 5 X3 k; J5 e( D2 T" I* E) g else StackPush(stack,0); " Y- y0 h! E6 b% R3 d if(a&0x02) StackPush(stack,1); 6 [6 s* |3 r3 F3 Q4 S2 I else StackPush(stack,0); `/ r" E4 h4 o2 S( S9 S if(a&0x04) StackPush(stack,1); / i: x3 @; `: @7 Y else StackPush(stack,0); / {5 N7 g8 b/ ?; W3 W( I' p0 z1 T if(a&0x08) StackPush(stack,1); # J' }$ D* R7 x2 \! W- s else StackPush(stack,0); . K/ s( a( E. T/ ~ }; W9 U, h$ y6 F1 k# q for(i = 0;i<8;i++)! b* w5 V- S" O. l$ n4 y {$ d) b/ K. a0 x3 }' j7 Q" [ int d = 0; 8 w( G) T: X) q3 P) R0 d StackPop(stack,&d); 4 d) m4 J. D6 i1 F( L if(i == 0)' B l9 _0 F1 y+ G5 K- { if(d == 1) b |= 0x80;0 o+ `" j3 Y- j5 n if(i == 1) 6 a/ S2 H) v* p0 n& C) R if(d == 1) b |= 0x40;5 ]+ B6 b+ F* C/ B; l ! g1 D f$ N1 D; T0 C+ } if(i == 2)5 E+ W1 l) ]* w: W# o, ^ if(d == 1) b |= 0x20;3 B/ h/ C1 \0 c/ R3 \& y6 c , e9 R+ ^+ {# ]8 s if(i == 3) , j B# L! D+ r& }9 g$ j if(d == 1) b |= 0x10; + m: e! y9 [) y$ v - s1 _8 a2 j1 h/ R if(i == 4)% S9 s9 [3 }8 k9 u9 \0 c7 b if(d == 1) b |= 0x08;: h( W( R6 r; M) S& N1 v 2 }) k8 g/ z) U8 z% `5 m if(i == 5) - y; Q K) q- c% r- R3 Y3 ~# H if(d == 1) b |= 0x04;8 j- S/ u2 f0 y $ b) S" B: x9 g0 c if(i == 6). G9 y' e3 r& N if(d == 1) b |= 0x02;" p$ i, Z% c' h) n! ~ . U; A3 [* t1 k/ I5 Z. S& v/ l2 z/ E0 [- k if(i == 7) 4 @1 U) g/ }- Q* ? if(d == 1) b |= 0x01;9 K" }+ y: \) J0 h , W% a# `: m5 A( P6 t7 W) |( J j }8 ^9 i+ j/ q7 m, q & m: L7 b' q" w7 c" U printf("\nb:0x%x\n",b);7 s7 ^; S5 G- O `& p5 W return 0; 8 S, k( K" @; `* o ]} # {9 |* ], {1 P " h4 [. P3 [) H2 `4 r% C2 [执行如下： ' \$ x( _8 d1 X! b# {( G 1 R7 s( H% C ]! W. I% K 除了用栈来实现，群里有位同学的答案让我眼前一亮: _8 @8 Y6 Q# S1 o! W# e- | #include / } b7 ~/ Z+ z" E! N! r, o3 s! r3 F struct charbits{ ^5 c. p% i7 v% |6 u C6 h# a unsigned char data1 : 4;, m' R @/ R( p unsigned char data2 : 4;8 s5 `! [4 p3 Q* M }; 7 W& Q/ E' o, _9 W2 u/ S9 H$ Z3 U$ i+ O) e ^ 9 U$ R6 Y, ]- I; J e union dataChange{ ' |' M8 I/ l9 V% v1 T$ ] struct charbits charBits; ) G! K$ g: T8 [. E9 r+ ^ unsigned char data;* W* w$ _( F7 H4 s4 t6 G7 R }datchange,*p; : u: f6 I; j+ x' i0 I, t + z- I( @! L5 o" n2 F& Y* X' `int main(){( @. \6 q7 c! ~, {: x + x: k, q0 F3 b* f; ^; n unsigned char temp;# A8 U) f% a1 a5 @/ \ p4 H / H9 H& V3 C8 b/ I3 |3 Q p = &datchange; " a. j5 n1 @3 }2 N9 s3 K- g" F0 |7 s6 |: [ datchange.data = 0x34; " N: T5 F9 ?' ^3 p2 m1 H & L. B; w! o. d: S0 Y. U temp = p->charBits.data1;0 w* T$ t( X3 t ; p* x" Z* ?+ N" @# r p->charBits.data1 = p->charBits.data2; 7 v( z3 W* C+ o# v1 @9 b8 Z 7 o1 `6 |* F$ @! Y$ L- G/ E p->charBits.data2 = temp; 5 q, J7 V% u6 t" Q A: a; i" J+ i5 z, ^% ?( n printf("data is %x", datchange.data);( u; t( |4 M$ q7 z4 P0 ^ ; R. T' x! I' q# D( v# _1 m return 0;5 z2 k% W* ]1 x }0 V F( F$ C. v/ [- r( d) k$ l$ Y6 V ; ?1 c& y: @7 k% N$ O6 @ 完美演示了位域、共用体、结构体、指针的几个知识点~6 j' a3 e, r9 X0 M4 D9 w 测试下位域的作用 , Y) i5 U B! o S9 _, U#include 3 ^7 U' z* b& S$ r9 `2 D5 |7 P: @# p. d# Z1 {8 Y1 s struct charbits{5 h4 `4 q4 N# R( s$ K) R- l unsigned char data1 : 1;6 A* s0 [; j5 P/ Y) W5 M5 N# t unsigned char data2 : 3; Z" f# N* R- i# \4 R9 k }; / E: j6 _4 @. y - t3 F3 Z; K2 k4 X8 d# x- ^- Q & i, m% [# b/ k7 ~; d" H: @% Fstruct charbits1{ - r; a% U" Z9 k3 O' {/ F" w6 T& ? unsigned char data1; , t( x' M6 }. [6 f- v0 U0 k unsigned char data2;' z# a3 Q/ P' O };8 I* ^5 L: s L, X' x6 e , b- {+ e* T6 k2 d+ Fint main(){ 9 y) w9 H) q, R! ]) ]6 q. v8 ~4 ^! H [/ [ struct charbits a; 1 I' H/ x) g) Z* ? struct charbits1 b; " E/ F+ ^* H# l( ^4 X. u5 G printf("%d %d\n",sizeof(a),sizeof(b)); ) A6 J3 I! Z; A4 K& b* n1 U% p : O# E9 I8 |8 M; c( }) [ /*赋值1看看*/3 ^4 D& f3 H+ M8 V; K1 T a.data1 = 1;; t2 D5 J9 p8 B- t, N printf("%d\n",a.data1); / C8 r# J& D5 P6 z6 w3 Z' o( D& ]! _3 ] /*赋值3看看*/ # h/ a" v* N ` a.data1 = 3;) F' `. q! N* {& m) @ printf("%d\n",a.data1); " i9 n- h F. P: C; w: c! H 6 c% ]0 R& C3 k" a /*赋值3看看*/) Y5 l9 s q! y a.data2 = 3; $ l; i% a+ h$ ]' B6 Z# Z% B printf("%d\n",a.data2); ) r' Z4 V6 e" `, t- ?( e' z5 M6 u7 d @ /*赋值7看看*/; v0 \$ @( e& T& _% h% X: Y a.data2 = 7;- s5 P. M, Q& _- z6 M- R( u printf("%d\n",a.data2); . n0 A( m1 _4 s - f* f; g% M6 C$ b, R, P ! v6 h$ O, U/ J: i2 W; K /*赋值8看看*/ ) T+ `4 I+ L s# X0 a a.data2 = 8;, B |" b+ ^0 T printf("%d\n",a.data2); ' g5 ~. A: ]0 j7 Z- e return 0; ; Q+ K% b6 ?5 @ x$ ]} 6 n$ L7 }. D& W B 8 R: C* @' K5 U5 s! U# ~data1只能存 0和1，如果存其他值的话就会溢出变成0。9 v0 }9 @* X/ X! N6 W8 X* e data2是 3bits ，只能存 0~7，我们存 8 进去，就溢出变成 0。! x8 b9 p* g) m' v2 E* T* a ! T) M! S; B8 i6 ?! {' s/ B : H/ ?7 w; i& w5 b0 i! n* D W

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gaosmile 发布时间：2020-5-13 19:29

技术帖
文章封面:	-
文章简介:	-

昨晚在微信群看到一个读者发的面试题目，从网上截图出来的，我百思不得其解，题目如图。

幸好，我学过栈

然后我写了个小程序

第一个方法比较笨，当我写完自己的代码后，看到有同学发了自己的代码，我赶紧就发了个红包，一个是为了鼓励大家多讨论问题，一个是为了赞扬这样的行为。

第一个方法就是取出每一个bit，压栈，然后再出栈，你想怎么排序都没有问题。

#include "stdio.h"
#include "stdlib.h"

struct List{
int data;
struct List * next;
};

struct Stack{
struct List *head;
int size;
};

struct Stack * StackInit(void)
{
struct Stack *stack = NULL;
stack =  (struct Stack*)malloc(sizeof(struct Stack));
stack->head = (struct List *)malloc(sizeof(struct List));
stack->head->next = NULL;
stack->size = 0;
return stack;
}

int StackPush(struct Stack *stack,int data)
{
struct List *tmp = (struct List *)malloc(sizeof(struct List));
tmp->data = data;
tmp->next = stack->head->next;
stack->head->next = tmp;
stack->size++;
//printf("push:%d \n",data);
return 0;
}

int IsStackEmpty(struct Stack *stack)
{
if(stack->head->next == NULL){
//printf("stack null\n");
return 1;
}else{
//printf("stack is not null\n");
return 0;
}
}

int StackPop(struct Stack *stack,int *data)
{
struct List *tmp = NULL;
if(IsStackEmpty(stack))
return -1;

tmp = stack->head->next;
*data = tmp->data;
stack->head->next = tmp->next;
stack->size--;
if(tmp!=NULL)
free(tmp);

//printf("pop:%d \n",*data);
return 0;
}

int main(void)
{
int i = 0;
unsigned char a = 0x71;
unsigned char b = 0;

struct Stack *stack = NULL;
stack = StackInit();

printf("a:0x%x\n",a);
for(i = 0;i<8;i++)
{

if(a&0x10) StackPush(stack,1);
else    StackPush(stack,0);
if(a&0x20) StackPush(stack,1);
else    StackPush(stack,0);
if(a&0x40) StackPush(stack,1);
else    StackPush(stack,0);
if(a&0x80) StackPush(stack,1);
else    StackPush(stack,0);

if(a&0x01) StackPush(stack,1);
else    StackPush(stack,0);
if(a&0x02) StackPush(stack,1);
else    StackPush(stack,0);
if(a&0x04) StackPush(stack,1);
else    StackPush(stack,0);
if(a&0x08) StackPush(stack,1);
else    StackPush(stack,0);
}
for(i = 0;i<8;i++)
{
int d = 0;
StackPop(stack,&d);
if(i == 0)
if(d == 1) b |= 0x80;
if(i == 1)
if(d == 1) b |= 0x40;

if(i == 2)
if(d == 1) b |= 0x20;

if(i == 3)
if(d == 1) b |= 0x10;

if(i == 4)
if(d == 1) b |= 0x08;

if(i == 5)
if(d == 1) b |= 0x04;

if(i == 6)
if(d == 1) b |= 0x02;

if(i == 7)
if(d == 1) b |= 0x01;

}

printf("\nb:0x%x\n",b);
return 0;
}

执行如下：

除了用栈来实现，群里有位同学的答案让我眼前一亮

#include <stdio.h>

struct charbits{
unsigned char data1 : 4;
unsigned char data2 : 4;
};

union dataChange{
struct charbits charBits;
unsigned char data;
}datchange,*p;

int main(){

unsigned char temp;

p = &datchange;

datchange.data = 0x34;

temp = p->charBits.data1;

p->charBits.data1 = p->charBits.data2;

p->charBits.data2 = temp;

printf("data is %x", datchange.data);

return 0;
}

完美演示了位域、共用体、结构体、指针的几个知识点~

测试下位域的作用

#include <stdio.h>

struct charbits{
unsigned char data1 : 1;
unsigned char data2 : 3;
};

struct charbits1{
unsigned char data1;
unsigned char data2;
};

int main(){

struct charbits a;
struct charbits1 b;
printf("%d %d\n",sizeof(a),sizeof(b));

/*赋值1看看*/
a.data1 = 1;
printf("%d\n",a.data1);
/*赋值3看看*/
a.data1 = 3;
printf("%d\n",a.data1);

/*赋值3看看*/
a.data2 = 3;
printf("%d\n",a.data2);

/*赋值7看看*/
a.data2 = 7;
printf("%d\n",a.data2);

/*赋值8看看*/
a.data2 = 8;
printf("%d\n",a.data2);
return 0;
}

data1只能存 0和1，如果存其他值的话就会溢出变成0。

data2是 3bits ，只能存 0~7，我们存 8 进去，就溢出变成 0。