How the printf() function works?

How the printf() function works?

I have wrote a simple program to use the c51's function printf().
as belows:

void main (void)
{
//initial_UART:
    SCON  = 0x50;        /* SCON: mode 1, 8-bit UART,
                      enable    rcvr      */
    TMOD |= 0x20;  /* TMOD: timer 1, mode 2, 8-bit
                    reload        */
    TH1   = 0xFD;  /* TH1:  reload value for 9600
                   baud @ 11.0592MHz   */
    TR1   = 1;     /* TR1:  timer 1 run */

    TI    = 1;     /* TI:   set TI to send first char
                     of UART    */
                  // TI is my question location!!
// main
        while(1)
        {
           printf("1234\n");
        }
}

When i set the TI=0 in the part of "//initial_UART", the printf() function will always
waiting for the the condition "TI=1", so ,it will not run out of the printf() function.
this can be see in the assembly mode: C:0x1234
JNB(TI(0x98,1),C:0x1234).

But when i set the TI=1 in the part of "//initial_UART" , this program will run well
and serialed out "1234" continually as i want.

My question is coming:
when the TI is set to 1 first, after the printf() run first time ,when "4" was sent out of SBUT ,the TI bit will be CLEAR by the printf() herself (the assemble line is : CLR TI(0x98,1)),when printf() want to run the second time ,it find the TI is 0,so printf() will always waiting for TI to 1 and halted, NOTHING will be serialed out i think, but the fact is the printf() function was run sucessfully again and again.

What's the matter? there is someting set the TI to 1 after send out the last data from the SBUF?
or the printf() have not RESET the TI to 0 when the last data send out ???But as we know,
TI should be reset to 0 when the TI was became to 1. am I right?