C99 printf 中的变量前导零 [英] Variable leading zeroes in C99 printf
问题描述
我正在用 C99 编写一个多精度库.根据编译我的代码的平台,我选择了不同的表示基础.
I'm writing a Multiprecision Library in C99. Depending on which platform my code is compiled I am selecting a different Base of representation.
例如,假设在平台 X 上,系统选择 BASE=100;并且在平台 Y BASE=10000 上;
So, for instance, let's say that on platform X the system select BASE=100; and on platform Y BASE=10000;
假设我代表 big unsigned int 如下:
Let's say I'm representing big unsigned int as follow:
typedef struct a {
big_enough_uint *digits;
unsigned int length;
unsigned int last;
} bigUint;
所以当我在 BASE-100 系统上时,我希望我的打印功能是
So when i'm on BASE-100 system I want my print function to be
void my_print(bigUint *A){
unsigned int i=0;
fprintf(stdout,"%d",A->digits[0]);
if(i!= A->last){
for(;i<=A->last;i++)
fprintf(stdout,"%02d",A->digits[i]);
}
printf(stdout,"\n");
}
在 BASE-10000 系统上,我希望它像这样
While on BASE-10000 systems I want it to be something like
void my_print(bigUint *A){
unsigned int i=0;
fprintf(stdout,"%d",A->digits[0]);
if(i!= A->last){
for(;i<=A->last;i++)
fprintf(stdout,"%04d",A->digits[i]);
}
printf(stdout,"\n");
}
我为什么要这样做??
假设我有以下号码:
12345600026789
在 BASE-100 表示中,数字数组将是(小端形式):
In BASE-100 representation the digits array will be (little-endian form):
12|34|56|0|2|67|89
^ ^ I want ONE LEADING ZEROES
在 BASE-10000 中:
while in BASE-10000:
12|3456|2|6789
^ I want THREE LEADING ZEROES
有没有简单的方法可以做到这一点?
Is there a simple way to do that?
推荐答案
阅读 man printf
中字段宽度的 *
占位符.
Read about the *
place holder for the field width in man printf
.
printf("%0*d", 3, 42);
给予
042
和
printf("% *s", 42, "alk");
给予
<39 spaces>alk
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