@文心快码type mismatch: cannot convert from integer to int 文心快码 在大多数编程语言中,integer 和int 是相同的概念,表示整数类型,因此不存在从 integer 到int 的转换问题。 在编程中,integer 和int 通常被视为同义词,都表示整数类型。这种类型用于存储整数值,如正数、负数和零。在大多数编程语言中,包括 ...
int qscore=Integer.parseInt(qs);将未知类型的值转化为int 型，需要用到Integer.parseInt()方法强制转化。另外你的逻辑也有问题。应该先判断ps 是否为空，若为空pscore=0; qs的判断也一样。你的代码应该这样写：int pscore;int qscore;if(ps.equals("")){ pscore=0;}else{ pscore=Integer...
-1是int类型，不是integer类型的，肯定会爆类型不匹配的错，这样就好了 private Integer id = Integer.valueOf(-1);去掉初始值-1
问Integer to Roman“不兼容的类型: int无法转换为布尔型[在MainClass.java中]”EN根据罗马数字的规则...
问Integer to Roman“不兼容的类型: int无法转换为布尔型[在MainClass.java中]”EN您的算法非常接近，...
*@params the {@codeString} to be converted to an * {@codeInteger}. *@exceptionNumberFormatException if the {@codeString} does not * contain a parsable integer. *@seejava.lang.Integer#parseInt(java.lang.String, int) */publicInteger(String s)throwsNumberFormatException {this.value = parseInt...
Integer id=new Integer(-1);-1是int类型，不是integer类型的，肯定会爆类型不匹配的错，这样就好了 类型
...but then you never cast it to an integer, so the input came back as a String, even though you meant to store an integer or something. So I'd check the call stack, find out what the line was that you're having trouble with, and just look at what that line might be trying ...
java.lang.intege javalanginteger报错,遇到java.lang.Integercannotbecasttojava.lang.Double问题时,我们可以将Integer类型先转成String类型,然后再转成Double类型,具体操作如下:Integeri=10;doubletmp=Double.valueOf(itoString());
charAt(0); if (firstChar < '0') { // Possible leading "+" or "-" if (firstChar == '-') { negative = true; limit = Integer.MIN_VALUE; } else if (firstChar != '+') throw NumberFormatException.forInputString(s); if (len == 1) // Cannot have lone "+" or "-" throw ...