CODEWITHSUNDEEP
cstringchars
Assaf
Assaf

Reputation: 1124

sizeof with arrays in c

My assignment is to print the binary value of a decimal number, and I want to control the size of the array as I understood I should do so my program would work in all the compilers.

I don't understand briefly the operator sizeof, but I would appriciate if you can explain where should I, and why, put the sizeof in my program:

void translate_dec_bin(char s[]){
    unsigned int decNum;
    char st[MAX_LEN] = { 0 };
    int j = 0;
    sizeof(decNum, 4);
    decNum = atoi(s);

    while (decNum > 0){
        st[j] = decNum % 2;
        decNum = decNum / 2;
        j++;
    }

    while (j >=0){
        printf("%d", st[j]);
        j--;
    }
    printf("\n");

}

My thought is that when I print the number, i.e in the code:

printf("%d", st[j]);

I should put the operator. Is it right?

Upvotes: 0

Views: 124

Answers (2)

IllusiveBrian
IllusiveBrian

Reputation: 3204

http://en.wikipedia.org/wiki/Sizeof

Sizeof is for measuring the byte-length of a datatype in C (and C++). So, if I were to write

size_t a = sizeof(int);

a will generally be equal to 4 (see Jonathan Leffler's comment). This is because a 32-bit integer requires 4 bytes of memory (32 bits/8 bits in a byte = 4).

Answering your question about portability, sizeof(int) should work on any compiler.

You might find this question useful: Is the size of C "int" 2 bytes or 4 bytes?

To set the size of your char array to the bit-size of an int, this should work:

const size_t intsize = sizeof(int) * 8;//sizeof returns size in bytes, so * 8 will give size in bits
char st[intsize] = { 0 };

Upvotes: 1

Sunil Bojanapally
Sunil Bojanapally

Reputation: 12658

sizeof is a unary operation, meaning it takes only one operand or argument.

Upvotes: 1

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