如何在C中解释此printf

Question

I am going through some code where I got stuck reading a printf statement:

printf("%*s%s\n", stat[type].dent, "", buf);

states[type].indent is a number which is equal to 4 and buf is buffer of 215 bytes with 0 strored into it as string but not sure "" and %*s is meant for.

Can anybody help me reading this printf statement?

Answer 1

The * in the format code "%*s" tells printf that the next argument is a field width followed by the normal argument (a string in this case).

It's the same as eg "%4s" in your case, but the width can be set in runtime.

So here stat[type].dent is the field width, and the empty string is the string. So this prints the empty string with 4 characters width (so you get four spaces) followed by whatever is in buf .

I recommend a good reference on printf where it's all documented.

Answer 2

The * means that the width of the argument to be printed is dependent on an (integer) argument that precedes it.

From cplusplus :

printf ("Width trick: %*d \n", 5, 10); // 5 is the width, 10 is the integer being printed

Answer 3

The %*s is a string conversion. The * part means that a width for the field will be supplied as a parameter (that's how stat[type].dent is used). So it's printing an empty string in some specified width (call it N). In other words, its leaves N blank spaces before printing out buf .

Answer 4

*将被stat[type].dent的值代替，该值是一个数字，称为最小字段宽度。

Answer 5

It is a precision, which specifies maximum number of bytes for string conversions.

Asterisk (*), uses an integer argument, which specifies the value (for precision) to be used.

To print a string for a variable length,specify printf("%*s", l, string) . The l is substituted for the asterisk.

Answer 6

"%*s%s\\n" is a format that contains 3 directives .

"%*s"

This directive is the string conversion specification. The * causes the field width, an int , to be determined from the the next parameter after the format. As OP mentions, this is the value of 4 from stat[type].dent . The s conversion specification then gets the next parameter ( "" ) and treats it a pointer to char * , which it is. Thus printf() prints out "" , pre-padding first with spaces, up to 4, as needed. 4 padding spaces are needed here as the string length of "" is 0. Net result: 4 spaces printed.

"%s"

This directive, also a string conversion specification, takes the next parameter, buf , assumes it is a char * , which it is, and prints its contents out. Unclear if OP means it is all 0, if so, then nothing is printed.

"\\n"

This final directive is simply a text, so "\\n" is printed.