Bitwise rotate left function

Question

I am trying to implement a rotate left function that rotates an integer x left by n bits

• Ex: rotateLeft(0x87654321,4) = 0x76543218
• Legal ops: ~ & ^ | + << >>

so far I have this:

int rotateLeft(int x, int n) {
  return ((x << n) | (x >> (32 - n)));
}

which I have realized to not work for signed integers..does anyone have any ideas as how to fix this?

so now I tried:

int rotateLeft(int x, int n) {
  return ((x << n) | ((x >> (32 + (~n + 1))) & 0x0f));
}

and receive the error:

ERROR: Test rotateLeft(-2147483648[0x80000000],1[0x1]) failed... ...Gives 15[0xf]. Should be 1[0x1]

Answer 1

int rotateLeft(int x, int n) {
  return (x << n) | (x >> (32 - n)) & ~((-1 >> n) << n);
}

UPDATE:(thanks a lot @George)

int rotateLeft(int x, int n) {
  return (x << n) | (x >> (32 - n)) & ~(-1 << n);
}

not use '-' version.

int rotateLeft(int x, int n) {
    return (x << n) | (x >> (0x1F & (32 + ~n + 1))) & ~(0xFFFFFFFF << n);
}

//test program
int main(void){
    printf("%x\n",rotateLeft(0x87654321,4));
    printf("%x\n",rotateLeft(0x87654321,8));
    printf("%x\n",rotateLeft(0x80000000,1));
    printf("%x\n",rotateLeft(0x78123456,4));
    printf("%x\n",rotateLeft(0xFFFFFFFF,4));
    return 0;
}
/* result : GCC 4.4.3 and Microsoft(R) 32-bit C 16.00.40219.01
76543218
65432187
1
81234567
ffffffff
*/