我可以为字符指针数组分配一个整数吗？

Question

int k=5;
char* result = (char *)malloc(100 * sizeof(char));
result[count] = k;

Considering the above code, if I print the contents of result[count] it prints smiley symbols. I tried like below,but of no use.

result[count]=(char)k;

Can anyone help me?

I am percepting that malloc(100*sizeof(char)) will create 100 blocks each of size of character contiguously. Please correct me if I am wrong?

Answer 1

I'll disregard all the fluff about arrays and malloc as it is irrelevant to the problem you describe. You seem to essentially be asking "What will this code print:"

char c = 5; 
printf("%c", c);

It will print the symbol 5 in your symbol table, most likely a non-printable character. When encountering non-printable characters, some implementations choose to print non-standard symbols, such as smileys.

If you want to print the number 5, you have to use a character literal:

char c = '5';
printf("%c", c);

or alternatively use poor style with "magic numbers":

char c = 53;   // don't write code like this
printf("%c", c);

Answer 2

It is a problem of character representation. Let's begin by the opposite first. A char can be promoted to an int , but if you do you get the representation of the character (its code ). Assuming you use ASCII:

char c = '5';
int i = c; // i is now 0x35 or 53 ASCII code for '5'

Your example is the opposite, 5 can be represented by a char, so result[count] = k is defined (even if you should have got a warning for possible truncation), but ASCII char for code 5 is control code ENQ and will be printed as ♣ on a windows system using code page 850 or 437.

A half portable (not specified, but is known to work on all common architectures) way to do the conversion (int) 5 -> (char) '5' is to remember that code '0' to '9' are consecutive (as are 'A' to 'Z' and 'a to 'z'), so I think that what you want is:

result[count] = '0' + k;

Answer 3

First, let's clean up your code a bit:

int k=5;
char* result = malloc(100);
result[count] = k;

sizeof(char) is always equal to 1 under any platform and you don't need to cast the return of malloc() in C.

This said, as long as your count variable contains a value between 0 and 99, you're not doing anything wrong. If you want to print the actual character '5', then you have to assign the ASCII value of that character (53), not just the number 5. Also, remember that if that array of chars has to be interpreted as a string, you need to terminate it with '\\0':

int k=5, i;
char* result = malloc(100);
for (i=0; i<98; i++) {
    result[i] = ' ';
}
result[98] = 53;
result[99] = '\0';
printf("%s\n", result);

Answer 4

The elements in the array you're trying to allocate are all the size of a char, 1 byte. An int is 4 bytes thus your issues. If you want to store ints in an array you could do:

int result[100];

Followed by storing the value, but honestly from the way your questioned is posed I don't know if that's actually what you're trying to do. Try rephrasing the post with what you're trying to accomplish.

Are you trying to store 5 as a character?