convert int counter to char in for loop in c

I can't seem to find an answer as to why I am not able to type cast a counting integer into a char type within a for loop. Below is a simple instance in which I have tried to declare the type case before and after the loop begins but have not gotten any where. Fundamentally, I don't understand why this is not allowed and can't, or am too stupid to find any information on this.

#include <stdio.h>

int
main(void)
int i;
//char j; <-If i declare here
//char j=(char)i; <-or here there is no profound affect

for(i=0;i<9;++i)
char j=(char)i;
printf("i=%d, j=%cn", i, j); <--Case A: j=%c, Case B: j=%d
return 0;


OUTPUT

Case A Output printf("%c",j) (j initialized as char & typecast as char)
i=0, j=
i=1, j=
i=2, j=
i=3, j=
i=4, j=
i=5, j=
i=6, j=
i=7, j=
i=8, j=

Case B Output printf("%d",j) (j initialized as char & typecast as char)
i=0, j=0
i=1, j=1 <- if typecast from int to char worked these values should not
i=2, j=2 <- show up when using printf("%d",j)
i=3, j=3
i=4, j=4
i=5, j=5
i=6, j=6
i=7, j=7
i=8, j=8


I was expecting a printed value of the the number as a char instead of blank values. There could be an issue with terminal or ascii values not being represented properly, but that does not answer the root of my question. ** Case B shows numeric values for the j column. However, these values were printed as using a double place holder. If my type case from double to char had worked, these values would be either the associated ascii values for the numbers or garbage.** I'm not looking for a way around this, just to understand what is going on when the code is compiled or in runtime. Thanks!

j

printf("i=%d, j=%dn", i, j);

%c

i

0

for

2 Answers
2

The Terminal has maybe a problem with the NUL character when displayed with %c in the formated string. I changed the Program a bit..

#include <stdio.h>
int main(void)
int i, x=123456789;
//char j;
char j=(char)i;
printf("int i: %dn",i);
printf("int x: %dn",x);
printf("char j: %dn",j);
printf("hex j: 0x%02xn",j);
for(i=32;i<128;++i)
char j=(char)i;
printf("i=%d, j=0x%02x, j= %d, j =%cn", i, j, j, j);

printf("char j: %cn",j);
// Problems is the output as char?
// Programmers Notepad Consol Specific!
printf("Terminal Stops before this in PN :Dn");
return 0;



I think the Program does what you want when printing with %d or %x. But %c makes problems.
I used Programmers Notepad. Other Programs may not have my problem :D
Importand is that Printable characters start at decimal 32 with space. See ASCII Printable. Before that it is normal to get no output. %c tries to convert the decimal value in a printable character according to the ASCII table. Also have a look on the C reference.The %d is for signed decimal Integer.

Because the number you are trying to convert is bigger than a char. Char can only be 1 character, hence you can't fit 9 digits/characters into a single char. This would need to be a char/string to do what you are trying.

Or try changing x=0->9 but not more than a single digit.

char

x

(char)7

'7'

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