"C语言练习4"版本比较

1. 编写一个函数find(s,c)，在给定字符串s中查找字符c，返回c在s中第一次出现时所在的地址。如果找不到，返回NULL。

   1 /*sample code*/
   2 char *find(char *s, int c) {
   3    for(;*s!='\0';s++)
   4       if(*s==c)
   5          return s;
   6    return NULL;
   7 }

int find(char s[], int c) {
   int i;
   for(i = 0; s[i]!='\0'; i++)
      s[i]!=c;
   return i;
   }
   return 0;
}

find(char s[], char c) 
{
   int i;
   for(i=0; s[i] != '\0';) {
      if(s[i] = c)
         return c;
      else i++;
      }
   return NULL;
}

void find(s, c) {
   int *p;
   while((s=getchar())!=EOF) {
      for(p=0; *p!=c; p++)
         if *p = c
            return p;
         else
            return NULL; 
   }
}

void find(char s[], c) {
   int i, j;
   for(i=0; i<j; i++) {
      s[i] = c;
      i++;
   }
   return NULL;
}

int find(char s[], int c, int i) {
   int c, i;
   char s[];
   for(i = 0; s[i] !='\0'; i++);
      if(i==c);
      return i;
      else
      return NULL
}

void find(char *s, char c) {
   for(; *s!='\0'; s++){
      if(*s==c)
         return s;
      else
         return NULL;
   }
}

int find(char s[], char c[]) {
   int i, j;
   for(i=0; i!='\0'; i++) {
      for(j=i; s[j]==c[k]; j++,k++);
      if(c[k] ='\0' && k>0)
         return i;
   }
   return NULL;
}

1. 写一个函数transpose(a)转置方阵a。a是一个N行N列的二维数组。N为常数。

   1 /*sample code*/
   2 int transpose(int a[N][N]) 
   3 {
   4    int i, j;
   5    for(i=0; i<N; i++) {
   6       for(j=i+1; j<N; j++) {
   7          int temp = a[i][j];
   8          a[i][j] = a[j][i];
   9          a[j][i] = temp;
  10       }
  11    }
  12 }

int transpose(int a[N][N]) {
   int i, j, temp;
   for(i = 0; i < N; i++)
      for(j = 0; j < N; j++) {
         temp = s[i][j];
         s[i][j] = s[j][i];
         s[j][i] = temp;
      }
}

int transpose(int a[N][N]) {
   int i, j, b;
   for(i = 0; i < N; i++)
      for(j = 0; j < i; j++) {
         b= a[i][j];
         a[i][j] = a[j][i];
         a[j][i] = b;
      }
}

int transpose(int a[N][N]) 
{
   int i, j;
   for(i=0; i<N-1; i++)
      for(j=0; j<N-1; j++)
         a[i][j] = a[j][i];
}

1. 写出表示复数（包含实部和虚部）的结构体complex。写出复数的加法运算add(c1, c2)，c1和c2为两个复数，add返回一个复数为c1与c2的和。

   1 /*sample code*/
   2 struct complex {
   3    double real;
   4    double imag;
   5 };
   6 
   7 struct complex add(struct complex c1, struct complex c2)
   8 {
   9    c1.real += c2.real;
  10    c1.imag += c2.imag;
  11    return c1;
  12 }

struct complex {
   int i;
   int j;
} c1, c2;

struct complex add(struct complex c1, struct complex c2)
{
   c1.i += c2.i;
   c1.j += c2.j;
   return c1;
}


struck complex{
   int a;
   int i;
} c1, c2;

struck complex add(struck complex c1, struck complex c2) 
{
   struck complex c3;
   c3->a = c1->a + c2->a;
   c3->i = c1->i + c2->i;
   return struck complex c3;
}

1. 给定strfilter函数，给定main函数，请写isnotalnum函数，使得程序能够删除字符串中非数字非英文字母的字符。

   1 /*sample code*/
   2 void strfilter(char s[], int (*fun)(int) ) {
   3    int i, j;
   4    for(i=j=0; s[i]!='\0'; i++)
   5       if( !(*fun)(s[i]) )
   6          s[j++] = s[i];
   7    s[j] = '\0';
   8 }
   9 main() {
  10    char str[]="Hello, 123";
  11    strfilter(str, isnotalnum);/*Hello123*/
  12    printf("%s\n", str);
  13 }

   1 int isnotalnum(int c)
   2 {
   3    return !isdigit(c) && !isalpha(c);
   4 }

void isnotalnum(char s[]) {
   int i, j;
   for(i=j=0; s[i] != '\0'; i++)
      if(isalnum(s[i]))
         s[j++] = s[i];
   s[j] = '\0';
}

int isnottalnum(int c)
{
   if(c>='0' || c=<'9')
      return 1;
   else if(c>='a' || c<='z')
      return 1;
   else if(c>='A' || c<=Z)
      return 1;
   else
      return -1;
}

1. 写一个程序，输出所有命令行参数中最大的一个。比如程序名字为max，输入命令max aa cc dd bb运行程序，程序输出dd。

   1 /*sample code*/
   2 #include <stdio.h>
   3 
   4 int main(int argc, char *argv[]) 
   5 {
   6    int i, max = 1;
   7    for(i = 2; i < argc; i++)
   8        if(strcmp(argv[i], argv[max])>0)
   9            max = i;
  10    if( max < argc)
  11        printf("%s", argv[i]);
  12 }

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[]) 
{
   int i, j;
   i = 1;
   while(i<argc) {
      for(j=i; j<argc;j++) {
         if(strcmp(argv[i], [j])<0)
            i = j;
            break;
      }
   return i;
   }
   printf("%s", argv[i]);
}

int main(int argc, char *argv[]) {
   if(argc > 1)
      qsort(v, 1, argc);
   printf("%s", argv[argc-1]);
   printf("\n");
   return 0;
}

int main(int argc, char *argv[]) {
   int i;
   char *t;
   if(argc!=1) {
      t = argv[1];
      for(i=2; i<argc; i++) {
         if(strcmp(t,argv[i])<0 )
            t = argv[i];
      }
      printf("%s", t);
   }
}

#define MAX 500
int main(int argc, char *argv[]) {
   char s[MAX];
   strcpy(s, argv[1]);
   int i;
   while(i=2; i<argc-1; i++)
      if(strcmp(s, argv[i]) <0)
         strcpy(s, argv[i]);
   if(argc != 1)
      printf("%s", s);
   return 0;
}