在 C 中将 int 列表转换为 char 列表时会发生什么？

Question

In the example below, I have attempted to convert a list of ASCII integer values (representing "a, b, c") from an integer list to a character list:

int main() {

   int myList[3] = {97, 98, 99}; // a, b, c in ASCII
   char * myCharList = (char*) myList;
   
   printf("%c\n", myCharList[0]);
   printf("%c\n", myCharList[1]);
   printf("%ld\n", strlen(myCharList));
   return 0;
}

Output:

a

1

Why does casting the int list create a character list that is only of length 1?

Edit:

I think I am starting to understand now. So as another example, if I changed the code like this:

#include <stdio.h>
#include <string.h>

int main() {

   int myList[3] = {255 + 101, 98, 99};
   char * myCharList = (char*) myList;

   printf("%c\n", myCharList[0]);
   printf("%c\n", myCharList[1]);
   printf("%ld\n", strlen(myCharList));
   return 0;
}

Then the bytes would be 100, 255, 0, 0. Then the output would be:

d

2

And because an integer is two bytes, this pointer can not be a list longer than length 2.

Answer 1

You did not convert an integer list to a character list.

myList is an array of int . However, when an array is used in most expressions ¹ , it is converted to a pointer to its first element.

So (char*) myList converts a pointer to myList[0] to a pointer to char . No conversion of the list is performed; only the pointer is converted.

The result is that myCharList points to the bytes of myList in memory.

In your system, int values are stored with the least significant byte first. Likely, they have four bytes, so the bytes representing the int 97 are 0, 0, 0, and 97, in order from most significant to least significant. So the bytes in memory, from the low address to the high address, are 97, 0, 0, 0.

When you apply strlen to this, it sees the 97 byte and then it sees a zero byte, which marks the end of the string. So strlen reports the length of this “string” in memory is 1.

To convert a list of integers to a list of character strings containing numerals for those integers, such as converting 97 to “97”, you must use routines like sprintf (which can format numbers like printf but write the results into a character array instead of to standard output) or write other source code to perform the conversion. And you must provide memory for where the characters will be written, by defining arrays large enough to hold the results or using malloc to allocate space.

Footnote

¹ When it is not the operand of sizeof or unary & and is not a string literal used to initialize an array.