如何使用malloc将int变量转换为C中的char * array？

Question

I'm doing a school project and this problem came up. by the way, i can't use library.

How to convert a int variable to char array? I have tried this but it didn't work, tried a lot of other things and even magic doesn't work...

char *r = malloc(sizeof(char*));
int i = 1;
r[counter++] = (char) i;

Can someone help me? Thank you for your time.

Answer 1

In your code, you should allocate for char size and not char * . Please try with this code segment

char *r = malloc(sizeof(char) * N); // Modified here
int i = 1;
r[counter++] = i & 0xff; // To store only the lower 8 bits.

Answer 2

You could also try this:

char *r = malloc(sizeof(char));
char *s = (char*)&i; 
r[counter++] = s[0];

This is an other funny way to proceed and it allows you to access the full int with: s[0] , s[1] , etc...

Answer 3

If you can't use the library, you can't use malloc .

But this will work:

int i = 0;
char *p = (char *)&i;
p[0] = 'H';
p[1] = 'i';
p[2] = '!';
p[3] = '\0';
printf("%s\n", p);

Assuming your int is 32bit or more (and your char is 8).

It then follows that if you have:

int i[100];

You can treat that as an array of char with a size equal to sizeof (i) . ie

int i[100];
int sz = sizeof(i);     // probably 400
char *p = (char *)i;    // p[0] to p[sz - 1] is valid.

Answer 4

Do you mind losing precision? A char is generally 8 bits and an int is generally more. Any int value over 255 is going to be converted to its modulo 255 - unless you want to convert the int into as many chars as is takes to hold an int.

Your title seems ot say that, but none of the answers give so far do.

If so, you need to declare an array of char which is sizeof(int) / sizeof(char) and loop that many times, moving i >> 8 into r[looop_var] . There is no need at all to malloc, unless your teacher told you to do so. In whch case, don't forget to handle malloc failing.

Let's say something like (I am coding this w/o compiling it, so beware)

int numChars = sizeof(int) / sizeof(char);
char charArry[numChard];      // or malloc() later, if you must (& check result)
int convertMe = 0x12345678;
int loopVar;

for (loopVar = 0; loopvar < numChars)
{
  charArry[loopVar ] = convertMe ;
  convertMe = convertMe  >> 8;
}

Answer 5

You can use a union instead. Assuming that sizeof int == 4 ,

typedef union {
    int i;
    char[4] cs;
} int_char;

int_char int_char_pun;
int_char_pun.i = 4;
for (int i = 0; i < 4; i++) {
    putchar(int_char_pun.cs[i]);
}

Be careful; int_char.cs usually won't be a null-terminated string, or it might be, but with length < 4.

Answer 6

if you don't want to include the math library:

unsigned long pow10(int n);

void main(){
    char test[6] = {0};
    unsigned int testint = 2410;
    char conversion_started = 0;
    int i=0,j=0;float k=0;
    for(i=sizeof(test);i>-1;i--){
        k=testint/pow10(i);
        if(k<1 && conversion_started==0) continue;
        if(k >= 0 && k < 10){
            test[j++]=k+0x30;
            testint = testint - (k * pow10(i));
            conversion_started=1;
        }
    }
    test[j]=0x00;
    printf("%s",test);
}

unsigned long pow10(int n){
    long r = 1;
    int q = 0;
    if(n==0) return 1;
    if(n>0){
        for(q=0;q<n;q++) r = r * 10;
        return r;
    }
}

NOTE: I didn't care much about the char array length, so you might better choose it wisely.

Answer 7

hmm... what is wrong with the code below

char *r = malloc(sizeof(char) * ARR_SIZE);
int i = 1;
sprintf(r,"%d",i);

printf("int %d converted int %s",i,r);

will it now work for you