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1、C语言习题答案第1章1 .选择题(1)c(2)B (3)D2.填空题(l)main(2)main(3)有穷性、确定性、有零个或多个输入、有一个或多个输出、有效性(4)顺序、分支和循环(5)自顶向下，逐步细化、模块化设计、结构化编码第2章1 .选择题B B C DA DDB B A B B DC B2、填空题(1)数字、字母、下划线(2)0 (3)4、8 (4)(a+b)*c/(a1 b)(5)6 0 (6)1 6(7)3 (8)6、4、2(9)0 (1 0)1 0、6 (1 1)5.5 0 0 0 0 0 (1 2)1 2、4(1 3)double(1 4)0 (1 5)1 6 (1 6)6.
2、63、编程题(1)main()p r int f(int:%5 dnA float :%5 dn/z“char:%5 dn“double:%5 dn/zlong:%5 dn，z,s izeof(int),s izeof(float),s izeof(char),s izeof(double),s izeof(long);(2)t t include t t define RA T 1.6 0 9 3 4main()float k=0.0;p r int f(inp ut t he km:)；s canf(f，&k);p r int f(z/nmile:%f/z,k*RA T);第3章1 .选择
3、题(1)(1 0)：DDC DC DC DB C2.解析题(1)x=1 7 0,x=L-I MU 1 7 0,x=L_|ML.|25 2,1 7 0X=1 7 0,X=1 7 0 L-l U M ,X=L-J L-J L-l1 7 0,x=%6 da=5 1 3.7 8 9 1 8 5,a=M M5 1 3.7 9,a=5 1 3.7 8 9 1 8 4 5 7,a=5 1 3.7 8 9 1 8 4 5 7(2)a=3 U b=7 x=8.5 U y=7 1.8 2cl=A Mc2=a3.编程题(1)main()(int x,y;s canf(%d%d,&x,&y);p r int f(商数
4、二%d,余数二 d，x/y,x%y);s ys t em(p aus e);(2)main()double x,y,z,avg;s canf(/z%lf%lf%lfz/,&x,&y,&z);avg=(x+y+z)/3;p r int f(.llf/z,avg,avg);s ys t em(p aus e);第4章1.选择题(1)(1 0)C C A A DC C A B D2.填空题(1)ch=，A，&ch二 Zch=ch 3 2x-ly=-l(2)x 2(3)a+b ca=b&a=c&a+c b&b+c aa=ba=cb=c(4)mar k/1 0k=lcas e 9(5)x 0c=x/1
5、0y!=-23.编程题(1)t t include main()(int x;p r int f(，zp leas e inp ut anumber:);s canf(%d,&x);if(x%2=0)p r int f(zzx is a evennumber );els e p r int f(x is a oddnumber );(2)t t include main()(int x,y;p r int f(/zp leas e inp ut anumber );s canf(d，&x);if(x=-5)p r int f(zzt he number iser r or );els e if
6、(x 0)p r int f(%d,y);)els ey=x-l;p r int f(%d,y);els e尸x;if(x=0)if(x 1 0)y=x+l;p r int f(级d,y);els e p r int f(t he number iser r or );(3)#include main()int a,m;p r int f(/zp leas e inp ut a number:);s canf(%d,&a);s wit ch(a/1 0)cas e 0:cas e 1:cas e 2:m=l;br eak;cas e 3:m=2;br eak;cas e 4:m=3;br ea
7、k;cas e 5:m=4;br eak;default:m=5;br eak;)p r int f(%d,m);)(4)t t include main()float p r ice,t ax;p r int f(/zp leas e inp ut t he p r ice ofp r oduct:);s canf(%f,fep r ice);if(p r ice 1 0 0 0 0)t ax=p r ice*。.0 5;els e if(p r ice 5 0 0 0)t ax=p r ice*。.0 3;els e if(p r ice 1 0 0 0)t ax=p r ice*。.0
8、2;els e t ax=0;p r int f(%f,t ax);(5)t t include main()float s cor e;p r int f(/zp leas e inp ut t he s cor e ofs t udent:);s canf(f,&s cor e);if(s cor e=8 5)p r int f(VERY G OOD);els e if(s cor e=6 0)p r int f(G OOD);els e p r int f(B A D)；)第5章1 .选择题(l)d(2)c(3)a(4)d(5)a(6)d(7)d(8)b(9)d(1 0)b(ll)c(1
9、 2)b(1 3)d(1 4)a(1 5)c2.填空题(1)=0 m=m/k k+(2)5 4 6(3)3*i-2(4)-=*二 (5)8 5 2(6)j+i%j=0 j=i(7)s um k s um=二k j-2(8)s=0P=1 j=i3.改错题(1)第一处改正：F or改为for第 二 处 改 正：ave=s um/4改为ave=s um/4.0(2)第一处改正：j=9第二处改正：m=1 0 0*i+1 0*j+k3.编程题(1)t t include main()(int s;float n,t,s um;t=l;s um=0;n=l;s=l.0;while(n=1 0 0)s um
10、=s um+t;n=n+l;s=s;t=s/n;p r int f(/zs um=%1 0.6 fn，s um);)(2)利用辗除法，直到b为。为止main()int p,r,n,m,t emp;p r int f(inp ut t wo int eger n,m:)；s canf(/z%d,%d，&n,&m);if(n 0)xl=(x2+l)*2;/*第一天的桃子数是第2天桃子数加1后的2倍*/x2=xl;day-;p r int f(/zt he t ot al is%dn，xl)；)(4)t t include main()int i;long int n=l;for (i=l;i=1
11、0;i+)(n=n*i;p r int f(i%5=0?z/%2d!=%T0 1 dn ：级2d!=%T0 1 d，i,n);)(5)main()(int m,s,i;for(m=2;m 1 0 0 0;m+)s=0;for(i=l;i m;i+)if(m%i)=0)s=s+i;if(s=m)p r int f(/z%d it s fact or s ar e ，m);for(i=l;i m;i+)if(m%i=0)p r int f(线d,，i);p r int f(n);#include main()(int n,i,j;scanf(%d,&n);for(i=l;i=n;i+)for(j=
12、l;j=n+i-l;j+)if(j=n-i)printf(/z );else printf(*);printf(n);)(7)#includemain()(float xO,xl,x2,fxO,fxl,fx2;do p r int f(/zEnt er xl&x2:)；s canf(/z%f,%f，&xl,&x2);fxl=xl*(2*xl1 4)*xl+3)-6;fx2=x2*(2*x2-4)*x2+3)-6;while(fxl*fx2 0);do x0=(xl+x2)/2;fxO=xO*(2*x0-4)*x0+3)-6;if(fxO fxl)=le-5);p r int f(x=%6.2f
13、n，xO);)第 6章1.选择题C C DC C A A DB D2.写出程序的运行结果(1)s um=6 (2)Max=9,r ow=l,col=0(3)01 02 1 03 2 10(4)-53、填空(1)i=j=k=0 、i 4 、j(2)s i、s 0,s 1 s t r,s 1 s 2,s t r4、编程题(1)#include main()(int a1 2,count=0,i;r andomize();for (i=0;i 1 2;i+)ai=5+r andom(1 1);for(i=0;i12;i+)(printf(%5d,a i);count+;if(count%4=0)pr
14、intf(n)getchO;(2)main()(int a5=2,23,43,77,33);int max,min,i;long sum=OL;int count=0;float av;max=min=a0;for(i=0;imax)max=ai;if(ai min)min=ai;)av=(float)sum/5;for(i=0;iav)count+;printf(max:%dnmin:%dnav:%fncount:%d，max,min,av,count);getchO ;)(3)main()(intscore 30 =61,62,63,64,65,66,67,68,69,70,71,72,7
15、3,74,75,76,77,78,79,80,81,85,89,90,33,44,55,91,92,93);inti,count 9 0=0,count 8 0=0,count 7 0=0,count6 0=0;for (i=0;i 3 0;i+)(p r int f(/zint p ut NO.%d s t udent ss cor e:，i+1);s canf(/z%d/z,s cor e+i);)for (i=0;i=9 0)(count 9 0+;count 8 0+;count 7 0+;count 6 0+;)els e if(s cor ei=8 0)(count 8 0+;co
16、unt 7 0+;count 6 0+;else if(scorei=70)count70+;count60+;)else if(scorei=60)(count60+;printf(0 60:%dn60 70:%dn70 80:%dn80 90:%dn90 10 0:%d,30-count60,count60-count70,count70-count80,count80-count90,count90);printf(z/n60 or higher:%dn70 orhigher:%dn80 or higher:%dn90 orhigher:%dnz/,count60,count70,cou
17、nt80,count90);getch();(4)1、main()char a 4 5;int i,j,k;for (i=0;i 4 5;i+)ai二 for(i=0;i 5;i+)for(j=0;j i;j+)a9*i+j=；for(k=0;k 5;k+)a9*i+j+k=*;)j=0;for(i=0;i(x y?x:y)(9)s t r =abdef(1 0)8,1 7(1 1)a=5a*a=25a*a*a=5 1 2(1 2)si*XTX XT*XP XT XT*#xix viz*Jx 1*TX TX XTX XTX XT*XTXfx 1*1*TX XIX XT*XTX#1*TX TX
18、XT*TXxix viz*Jx 1*TX TX XTX XTX XT*XTX#si*1 1*XT*TX XTX T Xj3、编程题(1)long act (long n)if(n=l|n=0)r et ur n I L;els ereturn n*act(nT);main()(int n=3;p rin tf(n!=%ld,act(n);getchO;)(2)void mmax(float a,int n);main()float a6;int i;fo r(i=l;i)flagul;i+;+while(flag&i=A&(a)=a&(a)=zD)main()char c;s canf(c,&
19、c);p r int f(z/%dn/z,I SA LPH A(c);)(2)/*习题 8-2-2*/#define SWA P(t,x,y)t w;w=x;x=y;y=w;main()double a=9 9.9 9,b=ll.1 1;SWA P(double,a,b);p r int f(%f,%fn/z,a,b);(3)/*习题 8-2-3 */Sinclude s t dio.ht t define MA X 8 0t t define C H A NG E 1main()char s t r MA X;int i;p r int f(I np ut t ext:n/z);get s(
20、s t r);#if(C H A NG E)for (i=0;i=，a，&s t r izs t r i=，A&s t r i，Z，)s t r i+=l;els eif(s t r i=二z|s t r i=，Z)s t r i-=25;)t t endifp r int f(out p ut:n%s，s t r);)(4)/*习题 8-2-4 */t t define LEA P_Y EA R(y)(y%4=0)&(y%1 0 0!=0)|(y%4 0 0=0)main()int year;p r int f(Z/I np ut year:/z);s canf(d,&year);if(LE
21、A P_Y EA R(year)p r int f(/z%d is a leapyear.n，year);els ep r int f(/z%d is not a leapyear.n，year);)(5)/*习题 8-2-5.h*/t t ifndef X8 _2_5 _Ht t define X8 _2_5 _H#def ine A VER(a,b,c)(a+b+c)/3.0Sendif/*习题 8-2-5 */#include X8 _2_5.h main()int a,b,c;p r int f(I np ut t hr ee int eger sa,b,c:);s canf(%d,%
22、d,%d，&a,&b,&c);p r int f(z/A ver ageis%f.n/z,A VER(a,b,c);第 9 章4.编程题(1)/*习题 9-4-1*/#include ttinclude main()char*strl20,*str220,*str3 20;char swap();printf(Input three line:nz/);gets(strl);gets(str2);gets(str3);if(strcmp(strl,str2)0)swap(strl,str2);if(strcmp(strl,str3)0)swap(strl,str3);if(strcmp(str
23、2,str3)0)swap(str2,str3);printf(Now,the order is:n)；printf(/z%sn%sn%sn/z,strl,str2,str3)char swap(char*pl,char*p2)/*交换两个字符串*/char*p20;strcpy(p,pl);strcpy(pl,p2);strcpy(p2,p);(2)/*习题 9-4-2*/main()int number20,n,m,i;printf(Z/How many numbers?);/*共有多少个数*/scanf(%d,&n);printf(Input%d numbers:n，n);/*输入n 个
24、数*/for(i=0;in;i+)scanf(/z%d/z,&number i);printf(/zHow many place you want tomove?);/*后移多少个位置*/s canf(%d,&m);move(number,n,m);/*调用move函数*/p r int f(/zNow,t hey ar e:nz/);for (i=0;i ar r ay;p-)*p=*(p-l);*ar r ay=ar r ay_end;m;if(m 0)move(ar r ay,n,m);/*递归调用，当循环次数in减至0时，停止调用*/(3)/*习题 9-4-3*/#include#in
25、cludettdefine TO TAL 7int mseek(char*str,char xstr,int n)int i;for(i=0;in;i+)if(strcmp(stri,xstr)=0)return 1;return 0;)main()char*nameTO TAL=Lining,Linshan,1lanyuan,Zhangqiang,Haipo,rangbing;char xname20;printf(enter a name:);gets(xname);if(mseek(name,xname,TO TAL)printf(Found!n);els ep r int f(Not
26、found!n);(4)/*习题 9-4-4 */#include t t include int fun(char *s t r,int num4)int i;for(i=0;i=a&*s t r 二 z*s t r =A&*s t r 二 Z)num0+;els e if(*s t r二=)numl+;els e if(*s t r =O&*s t r 二 9 )num2+;els enum3+;s t r+;t t define N 8 0main()int s t r ingN,n4,i;get s (s t r ing);fun(s t r ing,n);for(i=0;i 4;i+
27、)p r int f(级dt ,ni);(5)/*习题 9-4-5 */t t include main()int a5 5,*p,i,j;void change(int *p);p r int f(I np ut mat r ix:nzz);for (i=0;i 5;i+)/*输入矩阵*/for(j=0;j 5;j+)s canf(%d,&aij);p=&aO0;/*使P指向o行o列元素*/change(p);/*调用函数，实现交换*/p r int f(/zNow,mat r ix:n)；for (i=0;i 5;i+)/*输出已交换的矩阵*/for(j=0;j 5;j+)p r int
28、f(4 d，aij);p r int f(n);void change(int *p)/*交换函数*/int i,j,t emp;int *p max,*p min;p max=p;p min=p;for (i=0;i 5;i+)/*找最大值和最小值的地址,并赋给pmax,pmin*/for(j=0;j*(p+5*i+j)pmin=p+5*i+j;)t emp=*(p+1 2);/*将最大值换给中心元素*/*(p+1 2)=*pmax;*pmax=t emp;t emp=*p;/*将最小值换给左上角元素*/*p=*pmin;*pmin=t emp;pmin=p+l;for(i=0;i 5;i+
29、)/*找第二最小值的地址赋给pmin*/for(j=0;j*(p+5*i+j)pmin=p+5*i+j;t emp=*pmin;/*将第二最小值换给右上角元素*/*pmin=*(p+4);*(p+4)=temp;pmin=p+l;for(i=0;i5;i+)/*找第三最小值的地址赋给pmin*/for(j=0;j*(p+5*i+j)pmin=p+5*i+j;/*将第三最小值换给左下角元素*/temp=*pmin;*pmin=*(p+20);*(p+20)=temp;pmin=p+l;for(i=0;i5;i+)/*找第四最小值的地址赋给pminVfor(j=0;j*(p+5*i+j)pmin=
30、p+5*i+j;temp=*pmin;/*将第四最小值换给右下角元素*/*pmin=*(p+24);*(p+24)=temp;)(6)/*习题 9-4-6*/ttinclude main()void avsco(float*psco,float*pave);void avcourl(char*pcou,float*psco);void fali2(char course510,intnum4,float score45,floataver4);void good(char course510,intnum4,float score45,floataver4);int i,j,*pnum,num
31、4;floatscore45,aver4,*psco,*pave;char course510,*pcou;printf(Input course:n)；pcou=course 0;for(i=0;i5;i+)scanf(%s，coursei);printf(Input NO.and scoures:n);printf(N0.);for(i=0;i5;i+)printf(,%s，coursei);printf(n);psco=&score0 0;pnum=&num0;for(i=0;i4;i+)scanf(%d,pnum+i);for(j=0;j5;j+)scanf(,%fz，,psco+5*
32、i+j);pave=&aver0;printf(nn);avsco(psco,pave);/*求出每个学生的平均成绩*/avcourl(pcou,psco);/*求出第一门课的平均成绩*/printf(nn);fali2(pcou,pnum,psco,pave);/*找出2 门课不及格的学生*/printf(nn);good(pcou,pnum,psco,pave);/*找出成绩好的学生*/)void avsco(float psco,float*pave)/*求每个学生的干均成绩的函数*/int i,j;float sum,average;for(i=0;i4;i+)sum=0.0;for(
33、j=0;j5;j+)sum=sum+(*(psco+5*i+j);/*累计每个学生的各科成绩*/average=sum/5;/*计算平均成绩*/*(pave+i)=average;void avcourl(char*pcou,float*psco)/*第一门课的平均成绩的函数*/int i;float sum,average1;sum=0.0;for(i=0;i4;i+)sum=sum+(*(psco+5*i);/*累计每个学生的得分*/averagel=sum/4;/*计算平均成绩*/printf(course 1:%s,averagescore:%6.2f.n，pcou,averagel)
34、;void fali2(char course510,intnum4,float score45,floataver4)/*找两门以上课程不及格的学生的函数*/int i,j,k,label;print f(/，=S t u dent w h o isfail=n);print f(/z N O.)；for(i=0;i 5;i+)print f(1 0 s，cou rse i);print f(/z av erag e。);for(i=0;i 4;i+)label=0;for(j=0;j 5;j+)if(score i j)=2)print f(级5 d,nu m i);for(k=0;k 5
35、;k+)print f(/z%1 0.2 f，score i k);print f(zz%1 0.2 fn，av er i);v oid g ood(ch ar cou rse 5 1 0,intnum4,floataver4)score45,float/*找成绩优秀的学生(各门85分以上或平均90分以上)的函数*/int i,j,k,n;printf(/=S tudent whosescore is good=n);printf(z/NO.)；for(i=0;i5;i+)printf(%10 s，coursei);printf(average、。“);for(i=0;i4;i+)n=0;fo
36、r(j=0;j85.0)n+;if(n=5)|(averi=90)printf(%5d,numi);for(k=0;k5;k+)printf(/z%10.2f，scorei k);printf(/z%10.2fn，aver i);/*习题 9-4-7*/ttinclude double sigma(doub1e(*fn)(double),double 1,double u)double sum=0,d;for(d=l;du;d+=0.1)sum+=fn(d);return sum;void main()double sum;sum=sigma(sin,0.1,1.0);printf(/zsum
37、 of sin from 0.1 to 1.0is:%fn/z,sum);sum=sigma(cos,0.5,3.0);printf(sum of cos from 0.5 to 3.0is:%fn，su m);(8)/*习题 9-4-8 */main()int i;ch ar*mont h name(int n);print f(z/inpu t M ont h N o.:n);scanf(%d,&i);if(i%sn/z,i,mont h name(i);/*调用指针函数 mont h name()*/)ch ar*mont h name(int n)/*定义一个指针函数mont h na
38、me(),返回一个指向字符串的指针*/st at ic ch ar*name =/z I lleg alM ont h ,丁 J anu ary,“F ebru ary,M arch”,丁 T J u ne,J u ly,“O ct ober”,“D ecember;A pril,M ay，“A u g u st”,“S ept ember”,“N ov ember”,ret u rn(n 1 2)?name 0:name n);(9)/*习题 9-4-9 */t t inclu de t t inclu de main()v oid sort (ch ar*p 口)；int i;ch ar*
39、p 1 0,st r 1 0 2 0;for(i=0;i 1 0;i+)p i=st r i;/*将第 i 个字符串的首地址赋予指针数组P的第i个元素printf(/zInput 10 strings:n/z);for(i=0;i10;i+)scanf(s,pi);sort(p);printf(/zNow,the sequence is:n)；for(i=0;i10;i+)printf(sn,pi);void sort(char*p 口)int i,j;charfor(i=0;i9;i+)for(j=0;j0)temp=*(p+j);*(p+j)=*(p+j+1)；*(p+j+1)=temp;
40、(1 0)/*习题 9-4-1 0 */#inclu de t t define L I N E M A X 2 0/*定义字符串的最大长度*/main()v oid sort(ch ar*p);int i;ch ar*p,*pst r 5,st r 5 L I N E M A X;for(i=0;i 5;i+)pst r i=st r i;/*将 第i个字符串的首地址赋予指针数组pst r的第i个元素*/print f(I npu t 5 st ring s:n/z);for(i=0;i 5;i+)scanf(%s,pst r i);p=pst r;sort (p);print f(st r
41、ing s sort ed:n/z);for(i=0;i 5;i+)print f(级sn，pst r i);v oid sort(ch ar*p)/*冒泡法对5个字符串排序的函数*/int i,j;ch ar*t emp;for(i=0;i 5;i+)for(j=i+l;j 0)/*比较后交换字符串地址*/t emp=*(p+i);*(p+i)=*(p+j);*(p+j)=t emp;第1 0章3.编程题(1)t t inclu de struct data int year;int month;int day;)；main()struct data a;intmonthnum 12=31,
42、28,31,30,31,30,31,31,30,31,30,31);int i,sum=0;scanf(d%d%d，&a.year,&a,month,&a.day);for(i=0;ia.month-1;i+)sum+=monthnumi;sum+=a.day;if(a.year%4=0&a.year%10 0!=0|a.year%40 0=0)sum+=l;printf(/z%d 年%d 月 d 日 is the%dday”,a.year,a.month,a.day,sum);system(pause);(2)#include ttinclude struct study float mid
43、;float end;float avg;；main()struct study student;scanf(z/%f%f/z,&student.mid,&student,end);student.avg=(student,mid+student.end)/2;printf(average scoreis%fn/z,student,avg);system(pause);ttinclude ttinclude struct stu int num;float mid;float end;float avg;s3;main()struct stu*p;for(p=s;pnum),&(p-mid),
44、&(p-end);p-avg=(p-mid+p-end)/2;)for(p=s;pnum,p-mid,p-end,p-avg);syst em(pau se);(4)#inclu de st ru ct st u int nu m;ch ar name 2 0;int ag e;st ru ct st u *nex t;；v oid list (st ru ct st u *h ead)st ru ct st u *p;print f(/zT h e list records are:nz/);p=h ead;if(h ead!=N U L L)do print f(%dt%st%dn,p-n
45、u m,p-name,p-ag e);p=p-nex t;w h ile(p!=N U L L);elseprintf(Z/The list is null);struct stu*insert(struct stu*head,struct stu*stud)struct stu*pO,*pl,*p2;pl=head;/*pl指向链表第一个节点*/pO=stud;/*pO指向要插入的节点*/if（head二 二NULL）/*原链表是空表*/head=pO;pO-next=NULL;/*pO 作为头指针*/else while(pl!=NULL)&(pO-num=pl-num)p2=pl;pl=
46、pl-nex t;/*pl指针后移*/if(pl!=N U L L)if(h ead=pl)h ead=pO;/*插入链表开头*/else p2-nex t=p0;/*插入到p2节点之后*/pO-nex t=:pl;else p2-nex t=p0;pO-nex t=N U L L;/*插入到最后*/)ret u rn h ead;)main()st ru ct st u *new st u,*h ead;h ead=N U L L;int nu m;scanf(d,&nu m);w h ile(nu m!=0)newstu=(struct stu)malloc(sizeof(struct s
47、tu);newstu-num=num;scanf(%s，newstu-name);scanf(%d，&newstu-age);head=insert(head,newstu);scanf(%d,&num);)list(head);system(z/pause/z);(5)#include void partition(unsigned long int num)union a unsigned int part2;unsigned long int w;n,*p;p=&n;n.w=num;printf(/zlong int=%lxn/z,num);printf Clow part num二%O
48、 x,high partnu m=%O x n，p-part 0,p-part 1);main()u nsig ned long int x;x=0 x 2 345 6 7 8 9;part it ion(x);syst em(pau se);)第n章一、选择1.A2.B3.C4.D5.B6.C7.B8.C9.c1 0.c1 1.a1 2.b二、填空1 :bi.dat fp2:fopen(fname,w )ch3:rfg et c(fp)4:二 二 N U L Lflag=ls st rlen(s)-l=，n三、改错题1：第一处改为：！feof(fp)第一处改为：I d2:第一处改为：rew
49、ind(fp)第一处改为：I d四、编程题1:#inclu demain()(ch ar s 8 0;int a;F I L E *fp;if(fp=fopen(t est ，w)二 二 N U L L)print f(C annot open file.n)；ex it (1);)fscanf(st din,%s%d，s,&a);fprint f(fp,/z%s%d，s,a);fclose(fp);if(fp=fopen(t est ，r)=N U L L)print f(C annot open file.n)；ex it (1);fscanf(fp,s%d，s,&a);fpritf(st
50、dout,s%dn，s,a)fclose(fp);2:#includemain()(FILE*fpl,*fp2;char ch;fpl=fopen(z/f ilel.c,r );fp2=fopen(/zfile2.c,w);ch=fgetc(fpl);while(!feof(fpl)(putchar(ch);ch=fgetc(fpl);rewind(fp l);w h ile(!feof(fpl)fpu t c(fg et c(fpl),fp2);fclose(fpl);fclose(fp2);3:#inclu demain()(F I L E *fp;long posit ion;fp=fo