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programming-examples/c/Basic/Can the size of an array be...

44 lines
1.4 KiB
C

#include<stdio.h>
#include <malloc.h>
#include<conio.h>
int main(int argc, char** argv) {
char** new_argv;
int i;
/* Since argv[0] through argv[argc] are all valid, the
program needs to allocate room for argc+1 pointers. */
new_argv = (char**) calloc(argc+1, sizeof (char*));
/* or malloc((argc+1) * sizeof (char*)) */
printf("allocated room for %d pointers starting at %P\n", argc+1, new_argv);
/* now copy all the strings themselves
(argv[0] through argv[argc-1]) */
for (i = 0; i < argc; ++i) {
/* make room for '' at end, too */
new_argv[i] = (char*) malloc(strlen(argv[i]) + 1);
strcpy(new_argv[i], argv[i]);
printf("
allocated %d bytes for new_argv[%d] at %P, "
"
copied \"
%s\"
\n"
,
strlen(argv[i]) + 1, i, new_argv[i], new_argv[i]);
}
new_argv[argc] = NULL;
/* To deallocate everything, get rid of the strings (in any
order), then the array of pointers. If you free the array
of pointers first, you lose all reference to the copied
strings. */
for (i = 0; i < argc; ++i) {
free(new_argv[i]);
printf("
freed new_argv[%d] at %P\n"
, i, new_argv[i]);
argv[i] = NULL;
}
free(new_argv);
printf("
freed new_argv itself at %P\n"
, new_argv);
return 0;
}