How to see contents of char * since printf won't work?

Here is a snippet of my code:

#include <stdio.h> #include "uhash.h" #include <openssl/evp.h> char * hash(item a) { const char * str= a.k; int len= strlen(str); int md_len; unsigned char md_value[EVP_MAX_MD_SIZE]; /* Buff to store change result */ EVP_MD_CTX *mdctx; /* Digest data structure declaration */ const EVP_MD *md; OpenSSL_add_all_digests(); md = EVP_get_digestbyname("SHA256"); mdctx = EVP_MD_CTX_create(); EVP_DigestInit_ex(mdctx, md, NULL); EVP_DigestUpdate(mdctx, str, len); EVP_DigestFinal_ex(mdctx, md_value, &md_len); EVP_MD_CTX_destroy(mdctx); char md5str[33]; for(int i=0;i<md_len;++i) { sprintf(&md5str[i*2],"%02x",(unsigned int)md_value[i]); } printf("%s\n", md5str); return md5str; } int main(int argv, char **argc) { char *c; if (argv>0) { int i=0; int s=0; for(i=1;i<argv;i++) { s+=strlen(argc[i]); } c=(char *)(malloc(sizeof(char)*s*(argv-2)+1)); s=0; for(i=1;i<argv;i++) { char *t=c+s; memcpy(t,argc[i],strlen(argc[i])); if(i!=argv-1){ printf("%d\n", argv); t[strlen(argc[i])]=' '; s++; } s+=strlen(argc[i]); } *(c+s)='\0'; } printf("%s\n", c); item * kee= malloc(sizeof(item)); kee->k=c; kee->v=10; char *res= hash(*kee); fflush(stdout); if(res==NULL) printf("result is null..."); else printf("%s\n",res); fflush(stdout); } 

So the main function takes the arguments fine(tested) and passes it fine(tested) but the hash() function although computing the hash, does not either return the right value or I can't print the result in main. I've been trying to check for errors since yesterday but I'm not very smart so any help would be greatly appreciated!