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printf formatting example

Fausto Bartra

in the keil c51 users manual, page 295

I found this example about printf formatting (see below)

some of the formatting type fileds have a "b" letter after the "%" symbol.

is there an explanation of what this "b" character does?

thank you

Fausto Bartra

#include <stdio.h>
void tst_printf (void) {
char a;
int b;
long c;
unsigned char x;
unsigned int y;
unsigned long z;
float f,g;
char buf [] = "Test String";
char *p = buf;
a = 1;
b = 12365;
c = 0x7FFFFFFF;
x = 'A';
y = 54321;
z = 0x4A6F6E00;
f = 10.0;
g = 22.95;
printf ("char %bd int %d long %ld\n",a,b,c);
printf ("Uchar %bu Uint %u Ulong %lu\n",x,y,z);
printf ("xchar %bx xint %x xlong %lx\n",x,y,z);
printf ("String %s is at address %p\n",buf,p);
printf ("%f != %g\n", f, g);
printf ("%*f != %*g\n", 8, f, 8, g);
}

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  • +1

    The letter "b" is therefore a byte qualifier:

    printf("char %d",a);

    first converts the variable "a" into an integer and then outputs it, but

    printf("char %bd",a);

    does not convert the variable "a" into an integer and is therefore more suitable for a device that is very resource-limited. However, both examples have the same effect, namely the output of the char variable "a" on the standard output.

Reply
  • +1

    The letter "b" is therefore a byte qualifier:

    printf("char %d",a);

    first converts the variable "a" into an integer and then outputs it, but

    printf("char %bd",a);

    does not convert the variable "a" into an integer and is therefore more suitable for a device that is very resource-limited. However, both examples have the same effect, namely the output of the char variable "a" on the standard output.

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