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I am having trouble converting a int64_t to a char array and back. I don't know what is wrong with the code below, it makes complete logical sense to me. The code works for a as shown, but not the second number b which clearly falls into the range of int64_t. #include <stdio.h> #include <stdint.h> void int64ToChar(char mesg[], int64_t num) { for(int i = 0; i < 8; i++) mesg[i] = num >> (8-1-i)*8; } int64_t charTo64bitNum(char a[]) { int64_t n = 0; n = ((a[0] << 56) & 0xFF00000000000000U) | ((a[1] << 48) & 0x00FF000000000000U) | ((a[2] << 40) & 0x0000FF0000000000U) | ((a[3] << 32) & 0x000000FF00000000U) | ((a[4] << 24) & 0x00000000FF000000U) | ((a[5] << 16) & 0x0000000000FF0000U) | ((a[6] << 8) & 0x000000000000FF00U) | ( a[7] & 0x00000000000000FFU); return n; } int main(int argc, char *argv[]) { int64_t a = 123456789; char *aStr = new char[8]; int64ToChar(aStr, a); int64_t aNum = charTo64bitNum(aStr); printf("aNum = %lld\n",aNum); int64_t b = 51544720029426255; char *bStr = new char[8]; int64ToChar(bStr, b); int64_t bNum = charTo64bitNum(bStr); printf("bNum = %lld\n",bNum); return 0; } output is aNum = 123456789 bNum = 71777215744221775 The code also gives two warnings that I don't know how to get rid of. warning: integer constant is too large for ‘unsigned long’ type warning: left shift count >= width of type

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