A correct recursive strlen

#include <stdio.h>

/* A recursive strlen */ 
int slen(char *s) 
{ 
 return (*s) == '\0'? 0 : slen(s + 1) + 1; 
} 
int main() 
{ 
 printf("strlen is %d\n", slen("abcdefg")); 
}

遞迴 strlen

#include <stdio.h>

/* A recursive strlen */ 
int slen(char *s) 
{ 
 return (*s) == '\0'? 0 : slen(s++) + 1; 
} 
int main() 
{ 
 printf("strlen is %d\n", slen("abcdefg")); 
}

4 * 5 = 20

C:\SVN\book\C\blog>cat even-1.c
#include 

// 程式將 i 調整為偶數後再乘以 5

int main()
{
  int i = 3;

  // 檢驗 i 是否為奇數
  if (i % 2 == 1) // 沒失敗
    i++;

  i *= 5; // 變成偶數後再乘以 5
  printf("%d\n", i);
}

C:\SVN\book\C\blog>gcc even-1.c

C:\SVN\book\C\blog>a.exe
20

4 * 5 = 15

C:\SVN\book\C\blog>cat even.c
#include 

// 程式將 i 調整為偶數後再乘以 5

int main()
{
  int i = 3;

  // 檢驗 i 是否為奇數
  if (i % 2 == 1) // 成功
    i++;

  i *= 5; // 變成偶數後再乘以 5
  printf("%d\n", i);
}

C:\SVN\book\C\blog>gcc even.c

C:\SVN\book\C\blog>a.exe
15

1 & 3 != 0

int main()
{
  char i = 1;
  char j = 3;

  if (i & j)
    printf("yes.\n");
  else
    printf("no.\n");

  return 0;
}

Output

yes.

1 & 3 == 0

int main()
{
  char i = 1;
  char j;

  scanf("%d", &j);
  if (i & j)
    printf("yes.\n");
  else
    printf("no.\n");

  return 0;
}

Input

3

Output

no.

補救的機會

期末常遇到學生來求情，所以P老師準備了以下說詞。

問題：老師我因為種種原因導致成績不及格，有沒有補救的機會？
答案:
短版本：沒有
長版本：沒有。成績的計算是作業xx%，考試yy%，其他因素0%。其他因素包括外務太多，會被二一，時間管理不當等等。我會這樣說並不是想耍酷，而是因為我想盡量把成績計算單純化，公平化。否則要考慮的因素太多，不單純，也不公平。成績計算就是反映你的學習成效，最單純公平的辦法就是看作業和考試。這是我的信念 ，也是我對這個問題的答案。

Old memories never disappear, they just change.

int *bar(int t)
{
  int i = t;
  int *temp = &i;
  printf("temp is %d, (*temp) is %d\n", temp, *temp);
  return temp;
}

void foo(int a, int b, int c)
{
  int i = 40;
}

int main()
{
  int *a;
  a = bar(10);
  printf("a is %d, (*a) is %d \n", a, *a);
  foo(10, 20, 30);
  printf("a is %d, (*a) is %d \n", a, *a);
}

The output.

temp is 2686740, (*temp) is 10
a is 2686740, (*a) is 10
a is 2686740, (*a) is 40

Beware of typos

#include <stdio.h>
int main()
{
  int type = 10;
  int i =10;
  switch (type) {
  case 1:
    i = 0;
    printf("i = %d\n", i);
    break;
  case 2:
    i = 4;
    printf("i = %d\n", i);
    break;
  defualt:
    i = 5;
    printf("i = %d\n", i);
    break;
  }
  return 0;
}

No output!

Short circuit evaluation

計算 gcd 的函式. 注意 && 的 short circuit evaluation.

int gcd(int a, int b)
{
if (b)
while((a %= b) && (b %= a));
return a + b;
}

因果業報

寫 qsort 的人都知道, 在寫比較函式時, 第一件事就是把傳進來的兩個 const void * 指標轉型成為可以操作的型態, 例如 int *. 那麼我們怎麼知道應該要轉成什麼型態?
這時候就要問當初傳進 qsort 的緩衝區是什麼型態了.
比較通俗的說法就是, 種瓜得瓜, 種豆得豆.
佛家的說法就是因果業報. 大般涅槃經有云: 善男子，知善因生善果，惡因生惡果，遠離惡因。如此看來, 指標轉型只是因果業報的表現而已, 只是把之前的因具體還原為現在的果罷了.中間就算經歷無相的境界, 最後還是要回到果報的天理. 所謂佛力不敵業力, 此之謂也.

Never swap with yourself

#include <stdio.h>

#define SWAP(x, y) x ^= y ^= x ^= y

main()

{

  int i = 3;

  int j = 5;

  printf("%d\n", i);

  printf("%d\n", j);

  SWAP(i, j);

  printf("%d\n", i);

  printf("%d\n", j);

  SWAP(i, i);

  printf("%d\n", i);

}

The output.

3
5
5
3
0

A program that prints itself

#include <stdio.h>

#include <assert.h>

int main(void)

{

  FILE *fp;

  int c;

  fp = fopen(__FILE__, "r");

  assert(fp != NULL);

  while ((c = fgetc(fp)) != EOF)

    putchar(c);

  fclose(fp);

  return 0;

}

Subtraction as a comparator

This is why you should not use subtraction as a comparator.

http://www.cplusplus.com/reference/clibrary/cstdlib/qsort/

/* qsort example */
#include <stdio.h>
#include <stdlib.h>

int values[] = {-2147483640, 50, 100};

int compare (const void * a, const void * b)
{
  return ( *(int*)a - *(int*)b );
}

int main ()

{

  int n;

  qsort (values, 3, sizeof(int), compare);

  for (n = 0; n < 3 ; n++)

     printf ("%d ",values[n]);

  return 0;

}

The output is.

50 100 -2147483640

/ 2 and >> 1

#include <stdio.h>
int main()
{
  int i = -13;
  if ((i / 2) == (i >> 1))
    printf("yes\n");
  else
    printf("no\n");
  return 0;
}

The output is no.

A > 0 and A < -1

int main()
{
  unsigned int ui = 2147483647;
  if (ui + 1 > 0)
    printf("ui + 1 > 0\n"); 
  if (ui + 1 < -1)
    printf("ui + 1 < -1\n");
  return 0;
}

The output.

C:\SVN\book\C>a.exe
ui + 1 > 0
ui + 1 < -1

A < 0, B > 0, but B > A is wrong?

A < 0, B > 0, but B > A is wrong?

int main()
{
  int i = 2147483647;
  unsigned int ui = 2147483647;
  if (i + 1 < 0)     
    printf("i + 1 < 0\n");   
  if (ui + 1 > 0)
    printf("ui + 1 > 0\n");
  if (ui + 1 > i + 1)
    printf("ui + 1 > i + 1\n");
  return 0;
}

The output.

C:\SVN\book\C>a.exe
i + 1 < 0
ui + 1 > 0

100

main(){{}{}{}

   {;;;;;;}
 {          }
{            }
{            }
{            }
 {          }
   {;;;;;;}

   {;;;;;;}
 {          }
{            }
{            }
{            }
 {          }
   {;;;;;;}}

Cannot open that file

Cannot open that file.

int main()
{
  char filename[80];
  FILE *fp;
  printf("input file name: ");
  fgets(filename, 79, stdin);
  fp = fopen(filename, "r");
  assert(fp != NULL);
  fclose(fp);
}

C:\SVN\book\C>ls -l fgets.c
-rw-rw-rw-   1 user     group         225 Dec 29 17:23 fgets.c
C:\SVN\book\C>a.exe
input file name: fgets.c
Assertion failed: fp != NULL, file fgets.c, line 10

變數靈異事件

#include <stdio.h>
#include <string.h>
int main(void)
{
  char source[] = "This is a string.";
  int i = 5;
  char destination[4];
  strcpy(destination, source);
  printf("i is %d\n", i);
  printf("source is [%s]\n", source);
  printf("destination is [%s]\n", destination);
  return 0;
}

The output.

i is 544434464
source is [a string.]
destination is [This is a string.]