I am trying to print a value in an array, but cannot print the value as normal. Trying to learn up on macros using C89 only. Here is the code:
#include<stdarg.h>
#include <stdio.h>
int getValues(int, ...);
int *myArr;
int getValues(int num_args, ...) {
int val[num_args];
va_list ap;
int i;
va_start(ap, num_args);
for(i = 0; i < num_args; i++) {
val[i] = va_arg(ap, int);
}
myArr = val;
va_end(ap);
return *val;
}
int main(void) {
getValues(1,2,3,4);
for(int i = 0; i < sizeof(myArr); ++i){
printf("%d\n", myArr[i]);
}
printf("Values are %d\n", myArr[0]); // Want this to print 1
return 0;
}
The posted code has several problems, comments inlined below.
int getValues(int num_args, ...) {
int val[num_args]; // <--- variable-length arrays did not exist in C89, not until C99
/*...*/
myArr = val; // <--- saves address of local array 'val' into global `myArr`
/*...*/
} // <--- but `val` ceases to exist once the function returns
int main(void) {
getValues(1,2,3,4); // <--- missing first argument, presumably '4' for 'num_args'
for(int i = 0;
i < sizeof(myArr); // <--- sizeof(myArr) == sizeof(int*) is not the array count
++i){
printf("%d\n", myArr[i]); // <--- `myArr` points to an array which no longer exists
}
/*...*/
}
The following is a possible rewrite with just the minimal modifications to make the code work.
#include <malloc.h>
#include <stdlib.h>
#include <stdarg.h>
#include <stdio.h>
int *getValues(int, ...);
int *getValues(int num_args, ...) {
int *val = malloc(num_args * sizeof(int));
if(val == NULL) return NULL;
va_list ap;
va_start(ap, num_args);
int i;
for(i = 0; i < num_args; i++) {
val[i] = va_arg(ap, int);
}
va_end(ap);
return val;
}
int main(void) {
int *myArr = getValues(5, 1,2,3,4,5);
if(myArr == NULL) { abort(); } // error
for(int i = 0; i < 5; ++i){
printf("%d\n", myArr[i]);
}
free(myArr);
return 0;
}
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