Difference between revisions of "C оператори"

From Ilianko
 
Line 11: Line 11:
  
 
<code><pre>
 
<code><pre>
 +
#include <stdio>
 +
 
int main()
 
int main()
 
{
 
{
int a = 1, b = 2;
+
  int a = 1, b = 2;
 
 
a++;
+
  a++;
printf(" a = %i \n" , a);
+
  printf(" a = %i \n" , a);
printf(" a++ = %i \n" , a++);
+
  printf(" a++ = %i \n" , a++);
printf(" ++a = %i \n" , ++a);
+
  printf(" ++a = %i \n" , ++a);
 
 
 +
  a = 1; b = 2;
 +
  a = b++; //  <=>  a = b; b = b + 1;
 +
  printf("a = b++; b = %i \n" , b);
 +
  printf("a = b++; a = %i \n" , a);
 
 
a = 1; b = 2;
+
  a = ++b; //  <=>  b = b + 1; a = a;
a = b++; //  <=>  a = b; b = b + 1;
+
  printf("a = ++b; b = %i \n" , b);
printf("a = b++; b = %i \n" , b);
+
  printf("a = ++b; a = %i \n" , a);
printf("a = b++; a = %i \n" , a);
 
 
 
a = ++b; //  <=>  b = b + 1; a = a;
+
  return 0;
printf("a = ++b; b = %i \n" , b);
 
printf("a = ++b; a = %i \n" , a);
 
 
return 0;
 
 
}
 
}
 
</code></pre>
 
</code></pre>

Latest revision as of 09:34, 23 March 2011

Математически + - * / %

 int a = 1, b = 2;
 a + b  = 3;
 a - b = -1;
 a * b = 2;
 a / b = 0; 
 a % b = 1
 ++a <=> a = a + 1 ; --a <=> a = a - 1
#include <stdio>

int main()
{
  int a = 1, b = 2;
	
  a++;
  printf(" a = %i \n" , a);
  printf(" a++ = %i \n" , a++);
  printf(" ++a = %i \n" , ++a);
	
  a = 1; b = 2;
  a = b++; //  <=>  a = b; b = b + 1;
  printf("a = b++; b = %i \n" , b);
  printf("a = b++; a = %i \n" , a);
	
  a = ++b; //  <=>  b = b + 1; a = a;
  printf("a = ++b; b = %i \n" , b);
  printf("a = ++b; a = %i \n" , a);
	
  return 0;
}
</code>

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[Оператори в C и C++]