printf("%s%s","somestring"); // prints somestringsomestring
有没有办法做到这一点?
解决方法
printf("%1$s%1$s\n","hello");
在此示例中,1 $表示“使用第一个参数”.我们也多次使用这种语法,因此我们可以使用一次给定的参数.
Linux man page for printf
提供了更多细节:
The arguments must correspond properly (after type promotion) with the
conversion specifier. By default,the arguments are used in the order
given,where each ‘*’ and each conversion specifier asks for the next
argument (and it is an error if insufficiently many arguments are
given). One can also specify explicitly which argument is taken,at
each place where an argument is required,by writing “%m$” instead of
‘%’ and “m$” instead of ‘‘,where the decimal integer m denotes the
position in the argument list of the desired argument,indexed
starting from 1. Thus,06001
and
06002
are equivalent. The second style allows repeated references to the same argument. The C99 standard does not include the style using ‘$’,which comes from the Single UNIX Specification. If the style using ‘$’ is used,it must be used throughout for all conversions taking an argument and all width and precision arguments,but it may be mixed with “%%” formats which do not consume an argument. There may be no gaps in the numbers of arguments specified using ‘$’; for example,if arguments 1 and 3 are specified,argument 2 must also be specified somewhere in the format string.