printf的字符串,长度可变项 [英] printf string, variable length item
问题描述
#define SIZE 9
int number=5;
char letters[SIZE]; /* this wont be null-terminated */
...
char fmt_string[20];
sprintf(fmt_string, "%%d %%%ds", SIZE);
/* fmt_string = "%d %9d"... or it should be */
printf(fmt_string, number, letters);
有没有更好的方法来做到这一点?
Is there a better way to do this?
推荐答案
有没有必要建立一个特殊格式的字符串。 的printf
允许您使用一个参数(即precedes值),如果使用指定precision一个。*
作为格式标签precision。
There is no need to construct a special format string. printf
allows you to specify the precision using a parameter (that precedes the value) if you use a .*
as the precision in the format tag.
例如:
printf ("%d %.*s", number, SIZE, letters);
请注意:有和宽度之间的区别(这是一个最小字段宽度)precision(可提供的最大字符数要打印)。%* S
指定宽度%:S
指定precision。 (你也可以使用%*。*
但随后你需要两个参数,一个是宽度一为precision)
Note: there is a distinction between width (which is a minimum field width) and precision (which gives the maximum number of characters to be printed).
%*s
specifies the width, %.s
specifies the precision. (and you can also use %*.*
but then you need two parameters, one for the width one for the precision)
又见printf的手册页( 3人的printf
Linux下),尤其是现场的宽度和precision的部分:
See also the printf man page (man 3 printf
under Linux) and especially the sections on field width and precision:
而不是十进制数字字符串的人们可以写的*或* M $(对于某些
十进制整数M)来指定字段宽度在未来给出
参数,或者在第m个参数,分别为,必须是int类型。
[...]
Instead of a decimal digit string one may write "*" or "*m$" (for some decimal integer m) to specify that the field width is given in the next argument, or in the m-th argument, respectively, which must be of type int. [...]
而不是十进制数字字符串的人们可以写的*或* M $(对于某些
十进制整数M)来指定precision将在未来给
参数,或者在第m个参数,分别为,必须是int类型的。
Instead of a decimal digit string one may write "*" or "*m$" (for some decimal integer m) to specify that the precision is given in the next argument, or in the m-th argument, respectively, which must be of type int.
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