Principle of printf indefinite length parameter

One 、printf() The parameters of the function are indefinite

• printf() The function formats a string , Then print to the screen , and printf() The parameters of a function are uncertain (>=1)

• printf() Declaration of functions ：int printf(const char *format, …);

• Use printf Function formatting is similar to filling in the blanks , It's divided into two parts const char *format yes subject , ... yes Options , There is only one requirement for doing the right job ： According to the order of the questions , The variable of the option is the variable type specified by the topic

#include <stdio.h>

int main(int argc, char **argv)
{
    
    printf("%c say %s %d \n", 'I', "hello", 6);
    
    return 0;
}

Two 、printf() Realize the principle of variable length of parameters

• Parameter transfer is stored in Stack Passed in , Indefinite parameters are stored behind fixed parameters

• Indefinite parameters in memory , according to 4 or 8 Byte alignment （32 On the bit machine according to 4 Byte alignment ,64 On the bit machine according to 8 Byte alignment ）, For pointer variables , Store only the address pointed to by the pointer variable

• To write a simple version of my_printf(), Implement the above standards C Library functions printf() The effect of

#include <stdio.h>
#include <stdarg.h>
  
void my_printf(const char *format, ...)
{
    
	va_list arg;//va_list Is a macro , Is a pointer to a variable length parameter list , That is to say “ Options ” list 
	va_start(arg, format);// initialization arg, take format The last first address is assigned to arg
 
	while (*format)// Traverse " subject "
	{
    
		char ret = *format;// One char One char Ground traversal 
		if (ret == '%')
		{
    
			switch (*++format)// First ++ Then take 
			{
    
			case 'c':
			{
    
				char ch = va_arg(arg, int);//va_arg(arg,int), get arg Points to the value of the parameter , At the same time make arg Point to the next parameter ,int Used to name the current parameter type , Although the current parameter is char type , But with int, This is because the current parameter type is only used to get values from memory , and va_list Is in accordance with the 4 Byte or 8 Byte Alignment , Designated as char Will report a mistake 
				putchar(ch);
				break;						
			}
			case 's':
			{
    
				char *pc = va_arg(arg, char *);
				while (*pc)
				{
    
				    putchar(*pc);
				    pc++;
				}
				break;				
			}
            case 'd':
			{
               
				int in = va_arg(arg, int);
				putchar(in+'0');
				break;
			}
			default:
				break;
			}
		}
        
		else
		{
    
			putchar(*format);
		}
		format++;
	}
	va_end(arg);//va_end  In some implementations arg Change to invalid value , here , It's a arg The pointer points to  NULL, Avoid wild pointers  
}

int main(int argc, char **argv)
{
    
    my_printf("%c say %s %d \n", 'I', "hello", 6);
    
    return 0;
}