1、
int Copy(const char* source,const char* dest){
FILE *fSource,*fDest;
fSource=fopen(source,"r");
fDest=fopen(dest,"w");
if(fSource==NULL||fDest==NULL)
return 0;
char ch;
while(feof(fSource)==0){
fread(&ch,sizeof(char),1,fSource);
if(ch>=0x61&&ch<=0x7a)
ch-=0x20;
fwrite(&ch,sizeof(char),1,fDest);
}
fclose(fSource);
fclose(fDest);
return 1;
}
2、
int Count(const char* source,int* cWord,int* cDigit,int* cLetter,int* cAll){
(*cAll)=(*cWord)=(*cDigit)=(*cLetter)=0;
FILE *fSource;
fSource=fopen(source,"r");
if(fSource==NULL)
return 0;
char bch=0,ch=0;
while(feof(fSource)==0){
fread(&ch,sizeof(char),1,fSource);
(*cAll)++;
if(isalpha(ch))
(*cLetter)++;
else if(isdigit(ch))
(*cDigit)++;
else if(isspace(ch)||ispunct(ch)||iscntrl(ch)){
if(isalpha(bch))
(*cWord)++;
}
bch=ch;
}
fclose(fSource);
return 1;
}
3、
#include<stdio.h>
#include<stdlib.h>
int j,n,sumreal,sumimaginary;
typedef struct plurality
{
int real;
int imaginary;
struct plurality * next;
}pluralityNode;
void inputplurality(pluralityNode * plurality)
{
printf("please input the real part");scanf("%d",&plurality-> real);getchar();
printf("please input the imaginary part");scanf("%d",&plurality-> imaginary);
}
pluralityNode * creatList(int n)
{
pluralityNode * p, * plurality;
int i;
plurality=(pluralityNode *)malloc(sizeof(pluralityNode));
plurality->next=NULL;
for(i=n;i>0;--i)
{
p=(pluralityNode *) malloc(sizeof(pluralityNode));
inputplurality(p);
p->next=plurality->next;plurality->next=p;
}
return(plurality);
}
void printList(pluralityNode * plurality)
{
pluralityNode * current;
current=plurality;
while(current->next!=NULL)
{
for(j=0;j<n;j++)
sumreal=sumreal+current->real;
sumimaginary=sumimaginary+current->imaginary;
current=current->next;
}
printf("The sum is:%d+%di",sumreal,sumimaginary);
}
c语言中的word和byte是什么数据类型啊
word即“字”,一个字一般是16位二进制数;
byte即“字节”,一个字节是8位二进制数。
C语言中没有word、byte类型,char、short两种类型一般是8位,相当于byte型,int一般是16位,相当于word型吧。
转载请注明出处51数据库 » for?Cletter?Cword 求解几道c程序编程题
听说好名字都起不了了