在 C 中将 Char 数组转换为 Long

Question

This question may looks silly, but please guide me I have a function to convert long data to char array

void ConvertLongToChar(char *pSrc, char *pDest)
{
    pDest[0] = pSrc[0];
    pDest[1] = pSrc[1];
    pDest[2] = pSrc[2];
    pDest[3] = pSrc[3];
}

And I call the above function like this

long lTemp = (long) (fRxPower * 1000);
ConvertLongToChar ((char *)&lTemp, pBuffer);

Which works fine. I need a similar function to reverse the procedure. Convert char array to long. I cannot use atol or similar functions.

Answer 1

You can do:

union {
 unsigned char c[4];
 long l;
} conv;

conv.l = 0xABC;

and access c[0] c[1] c[2] c[3] . This is good as it wastes no memory and is very fast because there is no shifting or any assignment besides the initial one and it works both ways.

Answer 2

Leaving the burden of matching the endianness with your other function to you, here's one way:

unsigned long int l = pdest[0] | (pdest[1] << 8) | (pdest[2] << 16) | (pdest[3] << 24);

Just to be safe, here's the corresponding other direction:

unsigned char pdest[4];
unsigned long int l;
pdest[0] = l         & 0xFF;
pdest[1] = (l >>  8) & 0xFF;
pdest[2] = (l >> 16) & 0xFF;
pdest[3] = (l >> 24) & 0xFF;

Going from char[4] to long and back is entirely reversible; going from long to char[4] and back is reversible for values up to 2^32-1.

Note that all this is only well-defined for unsigned types.

(My example is little endian if you read pdest from left to right.)

Addendum: I'm also assuming that CHAR_BIT == 8 . In general, substitute multiples of 8 by multiples of CHAR_BIT in the code.

Answer 3

A simple way would be to use memcpy:

char * buffer = ...;
long l;
memcpy(&l, buff, sizeof(long));

That does not take endianness into account, however, so beware if you have to share data between multiple computers.

Answer 4

If you mean to treat sizeof (long) bytes memory as a single long, then you should do the below:

char char_arr[sizeof(long)];
long l;

memcpy (&l, char_arr, sizeof (long));

This thing can be done by pasting each bytes of the long using bit shifting ans pasting, like below.

l = 0;
l |= (char_arr[0]);
l |= (char_arr[1] << 8);
l |= (char_arr[2] << 16);
l |= (char_arr[3] << 24);

If you mean to convert "1234\0" string into 1234L then you should

l = strtol (char_arr, NULL, 10); /* to interpret the base as decimal */

Answer 5

Does this work:

#include<stdio.h>

long ConvertCharToLong(char *pSrc) {
    int i=1;
    long result = (int)pSrc[0] - '0';
    while(i<strlen(pSrc)){
         result = result * 10 + ((int)pSrc[i] - '0');
         ++i;
    }
    return result;
}


int main() {
    char* str = "34878";
    printf("The answer is %d",ConvertCharToLong(str));
    return 0;
}

Answer 6

This is dirty but it works:

unsigned char myCharArray[8];
// Put some data in myCharArray here...
long long integer = *((long long*) myCharArray);

Answer 7

char charArray[8]; //ideally, zero initialise
unsigned long long int combined = *(unsigned long long int *) &charArray[0];

Be wary of strings that are null terminated, as you will end up copying any bytes beyond the null terminator into combined ; thus in the above assignment, charArray needs to be fully zero-initialised for a "clean" conversion.

Answer 8

Just found this having tried more than one of the above to no avail:=(:

    char * vIn = "0";
    long vOut = strtol(vIn,NULL,10);

Worked perfectly for me. To give credit where it is due, this is where I found it: https://www.convertdatatypes.com/Convert-char-Array-to-long-in-C.html