为什么以下代码没有给出所需的答案？
昨天在一次采访中，面试官问了我一个问题：
为什么下面的代码没有给出想要的答案？
int a = 100000, b = 100000;
long int c = a * b ;
语言是 C。
我告诉面试官，我们首先将 100,000 * 100,000 算作 int（溢出），然后将其转换为 long。
Yesterday on an interview the interviewer asked me a question:
Why doesn't the following code give the desired answer?
int a = 100000, b = 100000;
long int c = a * b ;
The language is C.
I've told the interviewer that we count first the 100,000 * 100,000 as an int(overflow) and just then cast it to long.
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评论（3）
我猜测线索将是发生整数溢出，但由于值如此之低，我没有看到这种情况发生。
int（通常为 32 位）的最大（正）值为：2,147,483,647
计算结果为：100,000,000
更新：
更新后的问题：
100000 * 100000< /code> 而不是10000 * 10000会导致 10,000,000,000，这将导致发生溢出。然后将该值转换为 long。为了防止这种溢出，正确的方法是将乘法中的两个值之一转换为 long（通常是 64 位）。例如
（长）100000 * 100000I'm guessing the clue would be an integer overflow to occur, but with such low values, I don't see that happening.
Maximum (positive) value for int (usually 32bit) is: 2,147,483,647
The result of your calculation is: 100,000,000
UPDATE:
With your updated question:
100000 * 100000instead of10000 * 10000results in 10,000,000,000, which will cause an overflow to occur. This value is then cast to a long afterwards.To prevent such an overflow the correct approach would be to cast one of the two values in the multiplication to a long (usually 64bit). E.g.
(long)100000 * 100000这是因为它首先将其计算为 int，然后才将其转换为 long 变量。（因此它首先作为整数溢出，然后变成 long）
代码应该是
That's cause it first calculates it as an int, and only then casts it into a long variable.(so it first overflows as an integer and then becomes a long)
the code should be
100000*100000是10000000000(10,000,000,000)，它大于 32 位int可以表示的最大值 (< code>2,147,483,647），因此它溢出。a*b仍然是int，它不是long int，因为表达式a*b的成员都是int类型，因此它们不会转换为long int：此转换仅在a*b求值后才会发生，当结果被分配给c时。如果您希望a*b的结果为long int，则需要将至少一个操作数转换为long int：此外< code>long int 可以与
int具有相同的大小（也可以用 32 位表示）：这最有可能发生在 32 位应用程序中，通常，sizeof(int) == sizeof(long int) == 4。如果您需要
c为 64 位，您最好使用像 这样的变量int64_t，确保您是 64 位的。100000*100000is10000000000(10,000,000,000) which is greater than the maximum value a 32bitintcan represent (2,147,483,647), thus it overflows.a*bis still anint, it's not along int, since the members of expressiona*bare both of typeint, thus they aren't converted tolong int: this conversion will only happen aftera*bhas been evaluated, when the result is assignedc. If you want the result ofa*bto belong intyou need to convert at least one of the operands aslong int:Moreover
long intcould be of the same size ofint(it could be represented on 32 bits too): this is most likely to happen with 32 bit application where, usually,sizeof(int) == sizeof(long int) == 4.If you need
cto be of 64 bits you should better use a variable likeint64_t, that ensures you to be of 64 bits.