基于C语言利用哈夫曼树实现文件压缩的问题

char ch;
		while(feof(fp) == 0){
			fscanf(fp,"%c",&ch);
			str[ch]++;
		}
		printf("\n");
		for(i = 0; i < 1024; i++){
			if(str[i] != 0){
				count[n++] = i; 
			}
		}

 CreateHT(ht,node);
		char strarr[256][256];
		int aa,bb;
		for(aa=0; aa<256;aa++)
			for(bb=0;bb<256;bb++)
				strarr[aa][bb]=-1; 
		
		for(k = 0;k < node;k++){
			huffmanCode(ht,k,strarr[k]);
		}

void CreateHT(HTNode *ht, int n){
	int i, j, lnode, rnode n1, n2;
	for(i = n; i < 2*n-1; i++){
		lnode = 65432; rnode = lnode;
		n1 = 0; n2 = 0;
		for(j = 0; j < i; j++){
			if(ht[j].parent == -1){
				if(ht[j].weight <lnode){
					rnode = lnode;
					n2 =n1;
					lnode = ht[j].weight;
					n1 = j; 
				} else if(ht[j].weight < rnode){
					rnode = ht[j].weight;
					n2 = j; 
				}
			} 
		}
		ht[n1].parent = i;
		ht[n2].parent = i;
		
		ht[i].weight = lnode+rnode;
		ht[i].parent = -1;
		ht[i].lchild = n1;
		ht[i].rchild = n2;
	}
}

void replace(char oldstr[], char key[], char rep[]){
	int oldstrLen, lengthOfKey, lengthOfRep, i, j , flag;
	char tmp[1000]; 
	oldstrLen = strlen(oldstr);
	lengthOfKey = strlen(key);
	lengthOfRep = strlen(rep);
 
	for( i = 0; i <= oldstrLen - lengthOfKey; i++){
		flag = 1;
		for(j = 0; j < lengthOfKey; j++){
			if(oldstr[i + j] != key[j]){
				flag = 0; 
				break;
			}
		}
		if(flag){
			strcpy(tmp, oldstr);
			strcpy(&tmp[i], rep);
			strcpy(&tmp[i + lengthOfRep], &oldstr[i + lengthOfKey]);
			strcpy(oldstr, tmp);
			i += lengthOfRep - 1;
			oldstrLen = strlen(oldstr);
		}
	}
}

int retTen(char str[]){
	int ten = 0;
	int t = 0;
	for(t = 0;t < strlen(str);t++){
		ten += (str[t]-'0') * pow(2,strlen(str) - t - 1); 
	} 
	return ten;
}

int oldFileSize=getFileSize(path);
int newFileSize=getFileSize(Gopath);
		
printf("\n");
printf("压缩成功!,文件已压缩至%s\n",Gopath); 
printf("该文件压缩率为：%.2lf\n",(double)newFileSize/(double)oldFileSize);