Explanation of integer expression and integer expression in c ++

Explanation of integer expression and integer expression in c ++

Because we can't remember the range of numbers represented by short, int, long, and long, record them here.

The expression of int is related to the compiler. Generally, the compiler uses four bytes for representation. The four bytes have 32 bits. The first byte is used to indicate positive and negative values, and the other 31 bits are used to represent numerical values. The int value range is (-2 ^ 31,2 ^ 31-1) [Why can a negative number represent one more? in simple words, it is related to the original code, anti-code, and complement code. In the manual Supplementary Code, 1000 0000 indicates-128 ].

Other types of representation range:

Short: 2 bytes (-2 ^ 15 ~ 2 ^ 15-1) that is (-32768 ~ 32767), that is, (SHRT_MIN ~ SHRT_MAZ) (included in the header file );

Int: 4 bytes (-2 ^ 31,2 ^ 31-1) that is (-2147483648,2147483648), that is, (INT_MIN ~ INT_MAZ );

Long: 4 bytes (-2 ^ 31,2 ^ 31-1) that is (-2147483648,2147483648), that is (LONG_MIN ~ LONG_MAZ );

Long: 8 bytes (-2 ^ 63,2 ^ 63-1), that is, (-9223372042554775807,9223372010954775807), that is, (LLONG_MIN ~ LLONG_MAZ );

When the leetcode [204. Count Primes] is used to calculate the number of prime numbers smaller than the integer n, we can see an algorithm called the errado filter method. The general meaning is as follows: first mark the numbers 2 to n-1 as false, and then start traversing. When traversing to I, mark the number of multiples of I as true. When traversing to j, if it is not marked as true, it is a prime number.

I think the most liked SOLUTION IN THE Discuss OF leetcode is as follows:

int countPrimes(int n) {    if (n<=2) return 0;    vector<bool> passed(n, false);    int sum = 1;    int upper = sqrt(n);    for (int i=3; i<n; i+=2) {        if (!passed[i]) {            sum++;            //avoid overflow            if (i>upper) continue;            for (int j=i*i; j<n; j+=i) {                passed[j] = true;            }        }    }    return sum;}

However, I don't think this upper setting is necessary [Well, at that time, I didn't understand what its comment "// avoid overflow" refers ]. Because I think j = I * I, if it is too large, it will be greater than n directly without executing the loop. It has the same function as its continue. However, after I delete the upper sentence, an error is returned when the input is "499979. So I created a local project for debugging and found that there was an illegal access to the array. It was later found that the int indicates that the range is not enough. The size of sqrt (INT_MAX) is 46341. When I> 46341, the int cannot represent this big integer, then it will overflow into a negative number, so it will still be less than n, it will enter the loop, so accessing passed [j] will cause an error. There are two solutions:

1. Use a upper as above, so that you can use an integer;

2. Use the long type to represent j and I * I, that is, the loop is written as for (long j = (long) I * I; j