为什么此C程序不考虑for循环而输出“ 10”？

Question

#include <stdio.h>
int main()
{
  int i=10;
  for (int i=1;i<=20;i++)
    i++;
    printf("%d",i); 
  return 0;
}

I don't understand that why this C program always only prints the output as 10 irrespective of the condition that has been given in the for loop. I tried to execute it by making some changes in the for loop like for (int i=1;i<=50;i++) , but it returned the same output! What is the mistake here?

Answer 1

This is what your code actually says, with some comments to illustrate

#include <stdio.h>
int main()
{
    int i=10;               // Variable i is declared (=10)
    for (int i=1;i<=20;i++) // DIFFERENT variable i is declared, hiding the first variable (=1).
    {
        i++;                // The inner-i variable increments: 3, 5, 7, .... 19
    }                       // inner-i is complete, and disappears.  It is no longer active.

    printf("%d",i);         // The original Variable i is printed.
                            // It still has value 10, as originally initialized.
    return 0;
}

Answer 2

It is a for loop, so it increments according to what you write in its header : i++ here. You don't need to increment i manually in the loop body by doing i++ once more.

Further than that, if you declare a variable i before the loop, and write for (int i=1;i<=20;i++) , it declares another variable only for the scope of the loop. Since you didn't put any brackets around your loop code, the latter is only made of the line i++ . Then, the for scope ends, and you print the "other" i variable, which you declared earlier, and initialized at 10.

Answer 3

printf("%d",i); is not part of the for loop, only i++ is. Indentation is not significant in C, you need to use blocks to group operations together. Eg

for (int i=1;i<=20;i++) 
{
    i++;
    printf("%d",i); 
}

Answer 4

In C indentation is not part of the block-structure (like it is in Python).

That means your code is really something like this:

#include <stdio.h>
int main()
{
  int i=10;
  for (int i=1;i<=20;i++)
    i++;  // In loop

  printf("%d",i);  // Not in loop
  return 0;
}

Blocks of code have to be delimited with curly-braces { and } :

#include <stdio.h>
int main()
{
  int i=10;
  for (int i=1;i<=20;i++)
  {
    i++;  // In loop
    printf("%d",i);  // Also in loop
  }
  return 0;
}

Answer 5

C doesn't care about indentation (it only cares about whitespace to the extent that it separates tokens). If you want multiple statements to be part of a for loop (or while loop, or if statement, etc.) then you need delimit those statements with curly braces:

for ( i = 0; i < 10; i++ )
  printf( "this statement is part of the loop\n" );
  printf( "this statement is NOT part of the loop\n" );

vs.

for ( i = 0; i < 10; i++ )
{
  printf( "this statement is part of the loop\n" );
  printf( "so is this\n" );
}

It's generally a good idea to use curly braces even if you only have a single statement:

while ( some_condition )
{
  do_something();
}

You've also created two separate variables named i :

int i = 10; // first instance

for ( int i = 1; i <= 20; i++ ) // second instance of i "shadows" the first

This works because the for loop creates a new scope , and you can have variables in different scopes with the same name. The variable i in the for loop hides or "shadows" the variable i declared at the beginning of the program. When the for loop exits, the i declared within it ceases to exist, so you're printing the value of the i declared at the beginning of the program.

Answer 6

Indentation is not a necessary part to define a block of code in C.

According to this code, the 'i' that you increment is a different variable than what you want it to be.

The equivalent code is:

int main()
{
  int n=10;
  for(int i=1; i<=20;i++)   
    i++;
  printf("%d",n);
}

But the required code is:

int main()
{
  int i=10;
  for(int i=1; i<=20;i++)
  {
    i++;
    printf("%d",i);
  }
}