我想创建一个渐变背景，渐变在上半部分，在下半部分有一个纯色，如下图所示:

我不能，因为中心色展开覆盖底部和顶部。

我如何制作一个像第一张图片一样的背景?我怎么做一个小的中心色，不展开?

这是上面后台按钮的XML代码。

<shape xmlns:android="http://schemas.android.com/apk/res/android" android:shape="rectangle" >
    <gradient 
        android:startColor="#6586F0"
        android:centerColor="#D6D6D6"
        android:endColor="#4B6CD6"
        android:angle="90"/>
    <corners 
        android:radius="0dp"/>


</shape>

我正在查看这里的strlen代码，我想知道在代码中使用的优化是否真的需要?例如，为什么像下面这样的工作不一样好或更好?

unsigned long strlen(char s[]) {
    unsigned long i;
    for (i = 0; s[i] != '\0'; i++)
        continue;
    return i;
}

更简单的代码是不是更好和/或更容易编译器优化?

strlen在链接后面的页面上的代码是这样的:

/* Copyright (C) 1991, 1993, 1997, 2000, 2003 Free Software Foundation, Inc. This file is part of the GNU C Library. Written by Torbjorn Granlund (tege@sics.se), with help from Dan Sahlin (dan@sics.se); commentary by Jim Blandy (jimb@ai.mit.edu). The GNU C Library is free software; you can redistribute it and/or modify it under the terms of the GNU Lesser General Public License as published by the Free Software Foundation; either version 2.1 of the License, or (at your option) any later version. The GNU C Library is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU Lesser General Public License for more details. You should have received a copy of the GNU Lesser General Public License along with the GNU C Library; if not, write to the Free Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA. */ #include <string.h> #include <stdlib.h> #undef strlen /* Return the length of the null-terminated string STR. Scan for the null terminator quickly by testing four bytes at a time. */ size_t strlen (str) const char *str; { const char *char_ptr; const unsigned long int *longword_ptr; unsigned long int longword, magic_bits, himagic, lomagic; /* Handle the first few characters by reading one character at a time. Do this until CHAR_PTR is aligned on a longword boundary. */ for (char_ptr = str; ((unsigned long int) char_ptr & (sizeof (longword) - 1)) != 0; ++char_ptr) if (*char_ptr == '\0') return char_ptr - str; /* All these elucidatory comments refer to 4-byte longwords, but the theory applies equally well to 8-byte longwords. */ longword_ptr = (unsigned long int *) char_ptr; /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits the "holes." Note that there is a hole just to the left of each byte, with an extra at the end: bits: 01111110 11111110 11111110 11111111 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD The 1-bits make sure that carries propagate to the next 0-bit. The 0-bits provide holes for carries to fall into. */ magic_bits = 0x7efefeffL; himagic = 0x80808080L; lomagic = 0x01010101L; if (sizeof (longword) > 4) { /* 64-bit version of the magic. */ /* Do the shift in two steps to avoid a warning if long has 32 bits. */ magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL; himagic = ((himagic << 16) << 16) | himagic; lomagic = ((lomagic << 16) << 16) | lomagic; } if (sizeof (longword) > 8) abort (); /* Instead of the traditional loop which tests each character, we will test a longword at a time. The tricky part is testing if *any of the four* bytes in the longword in question are zero. */ for (;;) { /* We tentatively exit the loop if adding MAGIC_BITS to LONGWORD fails to change any of the hole bits of LONGWORD. 1) Is this safe? Will it catch all the zero bytes? Suppose there is a byte with all zeros. Any carry bits propagating from its left will fall into the hole at its least significant bit and stop. Since there will be no carry from its most significant bit, the LSB of the byte to the left will be unchanged, and the zero will be detected. 2) Is this worthwhile? Will it ignore everything except zero bytes? Suppose every byte of LONGWORD has a bit set somewhere. There will be a carry into bit 8. If bit 8 is set, this will carry into bit 16. If bit 8 is clear, one of bits 9-15 must be set, so there will be a carry into bit 16. Similarly, there will be a carry into bit 24. If one of bits 24-30 is set, there will be a carry into bit 31, so all of the hole bits will be changed. The one misfire occurs when bits 24-30 are clear and bit 31 is set; in this case, the hole at bit 31 is not changed. If we had access to the processor carry flag, we could close this loophole by putting the fourth hole at bit 32! So it ignores everything except 128's, when they're aligned properly. */ longword = *longword_ptr++; if ( #if 0 /* Add MAGIC_BITS to LONGWORD. */ (((longword + magic_bits) /* Set those bits that were unchanged by the addition. */ ^ ~longword) /* Look at only the hole bits. If any of the hole bits are unchanged, most likely one of the bytes was a zero. */ & ~magic_bits) #else ((longword - lomagic) & himagic) #endif != 0) { /* Which of the bytes was the zero? If none of them were, it was a misfire; continue the search. */ const char *cp = (const char *) (longword_ptr - 1); if (cp[0] == 0) return cp - str; if (cp[1] == 0) return cp - str + 1; if (cp[2] == 0) return cp - str + 2; if (cp[3] == 0) return cp - str + 3; if (sizeof (longword) > 4) { if (cp[4] == 0) return cp - str + 4; if (cp[5] == 0) return cp - str + 5; if (cp[6] == 0) return cp - str + 6; if (cp[7] == 0) return cp - str + 7; } } } } libc_hidden_builtin_def (strlen)

为什么这个版本运行得很快?

它是不是做了很多不必要的工作?

如何通过XPath提取属性节点的值?

一个示例XML文件是:

<parents name='Parents'>
  <Parent id='1' name='Parent_1'>
    <Children name='Children'>
      <child name='Child_2' id='2'>child2_Parent_1</child>
      <child name='Child_4' id='4'>child4_Parent_1</child>
      <child name='Child_1' id='3'>child1_Parent_1</child>
      <child name='Child_3' id='1'>child3_Parent_1</child>
    </Children>
  </Parent>
  <Parent id='2' name='Parent_2'>
    <Children name='Children'>
      <child name='Child_1' id='8'>child1_parent2</child>
      <child name='Child_2' id='7'>child2_parent2</child>
      <child name='Child_4' id='6'>child4_parent2</child>
      <child name='Child_3' id='5'>child3_parent2</child>
    </Children>
  </Parent>
</parents>

到目前为止，我有这个XPath字符串:

//Parent[@id='1']/Children/child[@name]  

它只返回子元素，但我希望有name属性的值。

对于我的示例XML文件，下面是我希望输出的内容:

Child_2
Child_4
Child_1
Child_3

在SQL Server 2008中截断日期时间值(如删除小时、分钟和秒)的最佳方法是什么?

例如:

declare @SomeDate datetime = '2009-05-28 16:30:22'
select trunc_date(@SomeDate)

-----------------------
2009-05-28 00:00:00.000

例如，我可能想:

tail -f logfile | grep org.springframework | <command to remove first N characters>

我在想tr可能有能力做到这一点，但我不确定。

P0137引入了函数模板std::launder，并在关于联合、生命期和指针的部分对标准做了很多很多修改。

这篇论文要解决的问题是什么?我必须注意语言的哪些变化?我们在洗什么?

我有一个小问题，XPath包含与dom4j…

假设我的XML是

<Home>
    <Addr>
        <Street>ABC</Street>
        <Number>5</Number>
        <Comment>BLAH BLAH BLAH <br/><br/>ABC</Comment>
    </Addr>
</Home>

假设我想找到文本中所有有ABC的节点，给定根元素…

所以我需要写的XPath是

/ * [contains(短信),‘ABC’)

然而，这不是dom4j返回的内容....这是dom4j的问题，还是我对XPath工作原理的理解，因为该查询只返回Street元素而不返回Comment元素?

DOM使Comment元素成为一个具有四个标记(两个)的复合元素

[Text = 'XYZ'][BR][BR][Text = 'ABC'] 

我假设查询仍然应该返回元素，因为它应该找到元素并在其上运行contains，但它没有……

下面的查询返回元素，但它返回的不仅仅是元素——它还返回父元素，这对问题来说是不可取的。

//*[contains(text(),'ABC')]

有人知道XPath查询只返回元素<Street/>和<Comment/>吗?

SQL中TRUNCATE和DELETE的区别是什么?

如果你的答案是针对特定平台的，请注明。

所以，如果我在我的主目录下，我想把foo.c移动到~/bar/baz/foo.c，但是这些目录不存在，有没有什么方法可以自动创建这些目录，这样你只需要输入

mv foo.c ~/bar/baz/ 

一切都会解决的吗?似乎可以将mv别名为一个简单的bash脚本，该脚本将检查这些目录是否存在，如果不存在，将调用mkdir，然后调用mv，但我想检查一下，看看是否有人有更好的主意。

XPath bookstore/book[1]选择书店下的第一个图书节点。

如何选择第一个匹配更复杂条件的节点，例如，第一个匹配/bookstore/book[@location='US']的节点