Logical shift with signed numbers

I am going through the Coursera course The Hardware/Software Interface and because it has already ended, I would not be able to get any feedback on the code.

The assignment is:


LogicalShift: shift x to the right by n, using a logical shift. We can assume that 0 ≤ n ≤ 31.

I am wondering if shifting by n and then back by 1 is a good solution.

int logicalShift(int x, int n) {
  int ba = 1>n; //shift the number
    int msbShifted = (a >> n) << 1; //shift the MSB by n -1 

    //if negative ^ the leading 1's to create a logical shift
    return msbShifted ^ numShifted;
}

The C standard does not guarantee that an int is 32 bits. Even though the problem guarantees that 0 ≤ n ≤ 31, your code fails if run on a machine with a 64-bit int. One way to resolve that problem is to be explicit about the size of your inputs:

#include 

int32_t logicalShift(int32_t x, int n) {
    …
}

The C standard says (C99 §6.5.7):


The result of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type or if E1 has a signed type and a nonnegative value, the value of the result is the integral part of the quotient of \$\dfrac{\mathrm{E1}}{2 ^ \mathrm{E2}}\$. If E1 has a signed type and a negative value, the resulting value is implementation-defined.

Therefore, your best bet is to treat x as an unsigned type:

int logicalShift(int x, int n) {
    return (unsigned int)x >> n;
}

StackExchange Code Review Q#65037, answer score: 12