我试图生成一个字符串,如果有[16,24 ..]列表和另一个列表[" 00"," 7F&#34 ; ..]
生成的列表是 [00000000000000000000000000000000,7F7F7F7F7F7F7F7F7F7F7F7F7F7F7F7F7F,......
在python中代码将是:
for l in range(16, 32 + 1, 8):
for b in ['00', '7f', '80', 'ff']:
do_something(b * l);
我不确定如何在生锈中实现这一点,到目前为止,我已经得到了:
for &i in [16u,24u,32u].iter() {
for n in ["00","7f","80","ff"].iter() {
}
}
但我不知道如何创建字符串。
答案 0 :(得分:4)
您可以使用std::str::StrAllocating::repeat
:
fn main() {
for &i in [16u, 24u, 32u].iter() {
for n in ["00", "7f", "80", "ff"].iter() {
println!("{}", n.repeat(i))
}
}
}
打印
00000000000000000000000000000000
7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f
80808080808080808080808080808080
ffffffffffffffffffffffffffffffff
000000000000000000000000000000000000000000000000
7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f
808080808080808080808080808080808080808080808080
ffffffffffffffffffffffffffffffffffffffffffffffff
0000000000000000000000000000000000000000000000000000000000000000
7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f7f
8080808080808080808080808080808080808080808080808080808080808080
ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff