adding a char at the end of a number

so, ive managed to add a number at the end of a char

Code:

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char chara[30] = "";
char charb[30] = "test ";


//charb = "testcom";


itoa(numa,chara,30);


strcat(charb,chara);


printf("%s\n",charb);

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now, I want to add a char at the end of a number, any help?

There is no such function itoa() in the Standard Library, there is a non-standard function in some compilers as an extension.

numa is not defined, or initialized.

You should look st the IBM web page for the Non-Standard atoi() function to see how it is used.

Also look at sprintf().

You should stick to Standard Library functions and avoid most of the extensions.

Work this out first.

Quote:

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Originally Posted by rstanley

There is no such function itoa() in the Standard Library, there is a non-standard function in some compilers as an extension.

numa is not defined, or initialized.

You should look st the IBM web page for the Non-Standard atoi() function to see how it is used.

Also look at sprintf().

You should stick to Standard Library functions and avoid most of the extensions.

Work this out first.

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Code:

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itoa(numa,chara,30);



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actually worked for me, it changed:

Code:

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numa = 1;

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into

Code:

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chara[30] = "1";

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From the IBM Website for the itoa() function:

Quote:

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The itoa() function coverts the integer n into a character string. The string is placed in the buffer passed, which must be large enough to hold the output. The radix values can be OCTAL, DECIMAL, or HEX. When the radix is DECIMAL, itoa() produces the same result as the following statement:

(void) sprintf(buffer, "%d", n);

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Try a value of numa for values other than '1', and see the results. I would think it would not be what you expect.

Please use the much simpler to use, sprintf() Standard Library function.

in fact, I've figured it out

Code:

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    char charx[30] = " xx";
    char chary[30] = ""; //for intx
    char charz; //not needed




    int intx;
    intx = 5;


    int inty; //not needed


    itoa(intx,chary,30);


    strcat(chary,charx);






    printf("%s", chary);

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Compare the versatility of sprintf() to multiple functions in your code:

Code:

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#include <stdio.h>

int main(void)
{
  int ival = 1234;
  double dval = 1.2345;
  char ary[64] = "";

// Build a string with an int, int in Hex, address of the int, and the the double

  sprintf(ary, "int=%d, hex=%x, addr=%p, double=%lf\n", ival, ival, (void *) &ival, dval);

  printf("ary = %s\n", ary);

  return 0;
}

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Code:

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Output:
ary = int=1234, hex=4d2, addr=0x7ffd3ba48224, double=1.234500

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