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Home Backend Development PHP Tutorial 再談PHP中的str_repeat函數(shù)兌現(xiàn)

再談PHP中的str_repeat函數(shù)兌現(xiàn)

Jun 13, 2016 pm 01:06 PM
input php str

再談PHP中的str_repeat函數(shù)實(shí)現(xiàn)

前段時間討論了在PHP中生成某多個某字符或字符串的字符串的問題,呵呵。這句話聽起來太別扭了。舉個例子就是,生成10個a的字符串就是aaaaaaaaaa。當(dāng)然這里的a可以是ab或任何其他字符串。

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在博客“生成固定長度的某字符的字符串 PHP ”中描述了一共5中方法。也對這五種方法的效率進(jìn)行了比對,其中以PHP中自帶的str_repeat函數(shù)效率最優(yōu),這也在意料和情理之中。

方法4,5都是對PHP本身的函數(shù)進(jìn)行了組配以實(shí)現(xiàn)相同的功能,雖然沒有str_repeat優(yōu)秀,但是其效率還是好過了方法1,和方法2.

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方法1,2是用PHP語言自己實(shí)現(xiàn)的邏輯,通過字符串的迭代拼接實(shí)現(xiàn)的,效率當(dāng)然不能和PHP中的builtin方法相提并論。

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今天看到了PHP中的str_repeat函數(shù)的實(shí)現(xiàn),仔細(xì)一看,原來和自己實(shí)現(xiàn)的方法1中的思路是“異曲同工”啊。那PHP中的str_repeat究竟是怎么做的呢?核心邏輯代碼如下:

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    result_len = input_len * mult;
    result = (char *)safe_emalloc(input_len, mult, 1);
    
    /* Heavy optimization for situations where input string is 1 byte long */
    if (input_len == 1) { 
        memset(result, *(input_str), mult); 
    } else {
        char *s, *e, *ee; 
        int l=0; 
        memcpy(result, input_str, input_len);
        s = result;
        e = result + input_len;
        ee = result + result_len;

        while (e<ee l="(e-s)" : memmove s e result>
<p>首先計(jì)算了結(jié)果字符串的長度。并分配了對應(yīng)的內(nèi)存空間。</p>
<p>然后將input_str拷貝到分配內(nèi)存空間的最開始,</p>
<p>然后迭代將已經(jīng)拷貝的內(nèi)容復(fù)制到剩余的內(nèi)存空間。</p>
<p>從代碼邏輯可以看出,復(fù)制的思路不是一份一份復(fù)制的,而是復(fù)制一份、兩份、四份、八份......</p>
<p>直到填滿所分配的內(nèi)存空間,然后獎內(nèi)存空間的最后一位賦值為'\0'。</p>
<p>?</p>
<p>初次看到這段代碼時對以下代碼邏輯有不解:</p>
<p>?</p>
<pre name="code" class="c">     while (e<ee l="(e-s)" : memmove s e>
<p>所分配的內(nèi)存空間一定是input_str的整數(shù)倍+1。因此如果每次只是向后移一倍原始字符串長度的話,只需要在剩余長度小于原始字符串長度時退出循環(huán),然后在將最后一位賦值為'\0'。不就可以了嗎?</p>
<p>?</p>
<p>如果你也是這么想的,那么你和我剛開始犯了同樣的錯誤。</p>
<p>因?yàn)?,原始字符串的?fù)制不是一份一份地進(jìn)行的。如果復(fù)制7份的話(即str_repeat("abc",7);),只需要復(fù)制4次(包括第一次)。</p>
<p>而且最后一次復(fù)制之前剩余的未賦值的內(nèi)存空間為原始字符串長度的3倍+1,即3*3+1。只需要將"abcabcabcabc"字符串中的前10個字符進(jìn)行復(fù)制。這樣,result_str最后一個字符必定是原始字符串的第一個字符,'a'。這個‘a(chǎn)’本來就是不要的。</p>
<p>?</p>
<p>索性直接:</p>
<pre name="code" class="c">result[result_len] = '\0';

其實(shí)就是最后字符串中那個不需要的'a'替換為字符串的結(jié)束符'\0'。

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這樣str_repeat函數(shù)就完成了。

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****************************************************************

此外:

博客“生成固定長度的某字符的字符串 PHP ”中的方法1,雖然思想上和str_repeat一致,但是采用了字符串拼接方式,而且每次拼接都要產(chǎn)生新的字符串,對內(nèi)存空間也是極大的浪費(fèi),如下示例:

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如果要生成8個‘a(chǎn)’的字符串,首先會生成'aa',然后是‘a(chǎn)aaa’,然后是‘a(chǎn)aaaaaaa’。循環(huán)次數(shù)和str_repeat一致,但是中間生成了'aa','aaaa'兩個無用字符串,而且,目標(biāo)字符串越長,則垃圾越多。而且,循環(huán)中采用字符串拼接方式,效率也不高。

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而str_repeat則沒有多余的無用空間的分配和釋放操作。

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