Creating a variadic function that prints any kind of format of ("c", "f", "i", "s"), but doesn't work

I've created a struct, that groups the format character and a pointer to the function which prints according to the formatter.

typedef struct formatter
{
char spec;
void (*print)(va_list);
} fmt;

Then, i've created functions that prints each format.

void print_char(va_list args)
{
printf("%c", va_arg(args, int));
}
void print_int(va_list args)
{
printf("%d", va_arg(args, int));
}
void print_float(va_list args)
{
printf("%f", va_arg(args, double));
}
void print_string(va_list args)
{
char *spec = va_arg(args, char *);
if (spec == NULL)
{
printf("(nil)");
return;
}
printf("%s", spec);
}

in the main variadic function i've created an array of the struct in order to loop over it.

void print_all(const char *const format, ...)
{
fmt f[] = {
{'c', print_char},
{'i', print_int},
{'f', print_float},
{'s', print_string},
{'\0', NULL}};
int i = 0, j;
char *separator = "";
va_list args;
va_start(args, format);
if (format == NULL)
{
return;
}
while (format[i] != '\0')
{
j = 0;
while (f[j].spec)
{
if (f[j].spec == format[i])
{
printf("%s", separator);
f[j].print(args);
separator = ", ";
break;
}
j++;
}
i++;
}
printf("\n");
va_end(args);
}

The problem came, when i compile the program and test the case below:

int main(void)
{
print_all("ceis", 'B', 3, "stSchool");
return (0);
}

It prints
B, 66,
Instead of printing
B, 3, stSchool

I need to know where the problem at.

I expect the problem becomes in va_arg in each function, but in order of the less detailed knowledge that i've of variadic functions, i can't change something.
and also i don't wanna implement it with the use of switch cases, in order to modularize my program.

If you pass a va_list parameter to a function that calls va_arg, the calling function is no longer allowed to use that va_list (other than calling va_end). You violate this rule because print_all calls other functions that call va_arg.

One way to fix this would be to put all the va_arg calls in the same function. But this wouldn't work well with your function pointer approach.

Another way to fix it is to pass va_list * to the helper functions instead of va_list. This is explicitly allowed by the standard, and you can then keep using the same va_list in other functions. So you'd need to modify all your helper functions (and the function pointer type) to take a pointer.

C standard reference (C17 7.16/3):

> The object ap [of type va_list] may be passed as an argument to another function; if that function invokes the va_arg macro
with parameter ap, the value of ap in the calling function is indeterminate and shall be passed to the
va_end macro prior to any further reference to ap.²⁵⁷

Footnote 257:

> It is permitted to create a pointer to a va_list and pass that pointer to another function, in which case the original
function may make further use of the original list after the other function returns.

Passing va_list by value to multiple function is invalid. You have to pass a pointer to va_list.

#include <stdio.h>
#include <stdarg.h>
typedef struct formatter {
char spec;
void (*print)(va_list*);
} fmt;
void print_char(va_list *args) {
printf("%c", va_arg(*args, int));
}
void print_int(va_list *args) {
printf("%d", va_arg(*args, int));
}
void print_float(va_list *args) {
printf("%f", va_arg(*args, double));
}
void print_string(va_list *args) {
char *spec = va_arg(*args, char *);
if (spec == NULL) {
printf("(nil)");
return;
}
printf("%s", spec);
}
void print_all(const char *const format, ...) {
fmt f[] = {
{'c', print_char},
{'i', print_int},
{'f', print_float},
{'s', print_string},
{'\0', NULL}};
const char *separator = "";
va_list args;
va_start(args, format);
if (format == NULL) {
return;
}
for (int i = 0; format[i] != '\0'; ++i) {
for (int j = 0; f[j].spec; f++) {
if (f[j].spec == format[i]) {
printf("%s", separator);
f[j].print(&args);
separator = ", ";
break;
}
}
}
printf("\n");
va_end(args);
}
int main(void) {
print_all("cis", 'B', 3, "stSchool");
}