字符数组的整数转换

Question

I've been trying to brush up on my C recently and was writing a program to manually parse through a PNG file.

I viewed the PNG file in a hex editor and noticed a stream of bytes that looked like

00 00 00 0D

in hex format.

This string supposedly represents a length that I am interested in.

I used getc(file) to pull in the bytes of the PNG file.

I created a char array as

char example[8];

to store the characters retrieved from getc .

Now, I have populated example and printing it with

printf("%#x, %#x, %#x, %#x", example[0]....

shows 0, 0, 0, 0xd which is exactly what I want.

However when I use

int x = atoi(example) 

or

int x = strtol(example, NULL, 16)

I get back zero in both cases (I was expecting 13). Am I missing something fundamental?

Answer 1

atoi converts strings like "0" to its numeric equivalent, in this case 0 . What you have instead is the string "\\0\\0\\0\\0\\0\\0\\0\\r" which is nowhere near numeric characters.

If you want to interpret your bytes as a number you could do something like

char example[4] = {0, 0, 0, 0xd};
printf("%d\n", *(uint32_t*) example);

You will notice (in case you're using a x86 CPU) that you will get 218103808 instead of 13 due to little endianness: the farther you go right the more significant the number gets.

As PNG uses big endian you can simply use be32toh (big endian to host endianess):

uint32_t* n = example;
printf("%u\n", be32toh(*n)

Answer 2

atoi and strtol expect text strings, while you have an array of binary values. To combine the individual bytes in an array to a larger integer, try something like:

uint32_t x = (a[0] << 24) | (a[1] << 16) | (a[2] << 8) | a[3];

Answer 3

atoi etc. operates on (ascii) strings.
You would get 123 for "123", which is in bytes 49 50 41 0.
What you have instead is binary 00 00 00 7B ... (well, endianess matters too).

Simple, but in this case wrong solution (ignoring endianess):
Cast the array address to int* and then get a value with * .

As integers in PNG are supposed to be big endian in any case,
the pointer casting would only work with big endian machines.
As portable solution, shifting the bytes with 24,16,8,0 and binary-or´ing them will do.