問題描述

我通常使用以下成語來檢查字符串是否可以轉換為整數.

I normally use the following idiom to check if a String can be converted to an integer.

public boolean isInteger( String input ) {
    try {
        Integer.parseInt( input );
        return true;
    }
    catch( Exception e ) {
        return false;
    }
}

只有我一個人，還是這看起來有點駭人聽聞?有什么更好的方法?

Is it just me, or does this seem a bit hackish? What's a better way?

查看我的答案(帶有基準，基于 早期答案.com/users/28278/codingwithspike">CodingWithSpike) 了解我為什么改變立場并接受 Jonas Klemming 的回答 這個問題.我認為這個原始代碼會被大多數人使用，因為它實現起來更快，更易于維護，但在提供非整數數據時速度會慢幾個數量級.

See my answer (with benchmarks, based on the earlier answer by CodingWithSpike) to see why I've reversed my position and accepted Jonas Klemming's answer to this problem. I think this original code will be used by most people because it's quicker to implement, and more maintainable, but it's orders of magnitude slower when non-integer data is provided.

推薦答案

如果您不關心潛在的溢出問題，此函數的執行速度將比使用 Integer.parseInt() 快 20-30 倍.

If you are not concerned with potential overflow problems this function will perform about 20-30 times faster than using Integer.parseInt().

public static boolean isInteger(String str) {
    if (str == null) {
        return false;
    }
    int length = str.length();
    if (length == 0) {
        return false;
    }
    int i = 0;
    if (str.charAt(0) == '-') {
        if (length == 1) {
            return false;
        }
        i = 1;
    }
    for (; i < length; i++) {
        char c = str.charAt(i);
        if (c < '0' || c > '9') {
            return false;
        }
    }
    return true;
}

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