我在add2_recur函数中遇到了问题。我想在字符串中添加一个字符数字。但我不知道如何将字符串返回到我的main函数，因此我可以打印出结果。我尝试使用函数指针，但我只返回字符串的第一个值。
有关如何执行此操作的任何建议都会有所帮助。
//check if string is valid
int digcheck_helper(char *theno, int start, int length) {
int charToInt = *(theno+start);
if(!((charToInt >= 48) &&(charToInt <= 57)))
return 0;
if(length == 0)
return 1;
return digcheck_helper(theno,start+1,length-1);
}
int digcheck(char *str, int start, int length) {
return digcheck_helper(string,start,length);
}
/**********************
****add recursive function**/
void add2_recur(char *num1, char *num2, int start, int carryDig) {
int singleChar1 = *(num1 + start), singleChar2 = *(num2 + start);
char *str = (char*) malloc(strlen(num1) + 2);
sum = singleChar1 + singleChar - 96;
if(carryDig == 1)
sum = sum + 1;
if(start < strlen(num1)) {
if(sum >= 10) {
sum = sum - 10;
str[start] = sum + 48;
carryDig = 1;
printf("sum of single digit is: %c\n", str[start]);
}
else if( sum < 10) {
str[start] = sum + 48;
carryDig = 0;
printf("sum of single digit is: %c\n", str[start]);
}
add2_recur(num1,num2,start+1,carryDig);
}
else if ((start == strlen(num1)) && (carryDig ==1)){
str[start+1] = 48;
printf("sum of single digit is: %c\n", str[start+1]);
}
}
void add2(char *n1, char *n2) {
add2_recur(n1,n2,0,0)
}
/*******************/
int main() {
char string1[20000], string2[20000], revStr1[20000], revStr2[20000];
int digit_1, digit_2, i;
printf("Enter first number >");
fgets(string1,20000,stdin);
string1[strlen(string1)-1] = '\0';
digit_1 = digcheck(string1,0,strlen(string1)-1);
//Check if string is valid integer
if(digit_1 = 0)
printf("This number is invalid\n");
else{
printf("Enter second number >");
fgets(string2,2000,stdin);
string2[strlen(string2)-1] = '\0';
digit_2 = digcheck(string2,0,strlen(string2-1);
if(digit_2 == 0)
printf("This number is invalid\n");
else
printf("1st num is %s\n2st num is %s\n", string1, string2);
}
// reverse string
for(i=0;i<strlen(string1);i++)
revStr1[i] = string1[(strlen(string1)-1) - i];
for(i=0;i,strlen(string2);i++)
revStr2[i] = string2[(strlen(string2) -1) - i];
// compare string and pass to add2
if(strlen(revStr1) < strlen(revStr2)) {
for(i = strlen(revStr1); i < strlen(revStr2); i++)
revStr1[i] = '0';
add2(revStr1,revStr2);
}
else if(strlen(revStr2) < strlen(revStr1)) {
for(i = strlen(revStr2); i < strlen(revStr1); i++)
revStr2[i] = '0';
add2(revStr1,revStr2);
}
else
add2(revStr1,revStr2);
return 0;
}
答案 0 :(得分：0)
this function:
void add2(char *n1, char *n2)
{
add2_recur(n1,n2,0,0)
}
this function has a couple of problems.
1) it will not compile because the statement that calls add2-recur()
is missing a trailing ';'
2) this function is not needed as add2_recur can be called directly
3) this is expected to add two numbers together ..
How is it to return the result?
It (probably) should be more like:
void add2(char *n1, char *n2, char *sum)
{
strcpy(sum, add2_recur(n1,n2,0,0) );
}
答案 1 :(得分：0)
//检查字符串是否有效
int digcheck_helper（char * theno，int start，int length）{
int charToInt = *(theno+start);
if(!((charToInt >= 48) &&(charToInt <= 57)))
return 0;
if(length == 0)
return 1;
return digcheck_helper(theno,start+1,length-1);
}
int digcheck（char * str，int start，int length）{
return digcheck_helper(string,start,length);
}
/ ********************** ****添加递归函数** /
char * add2_recur（char * num1，char * num2，int start，int carryDig）{
int singleChar1 = *(num1 + start), singleChar2 = *(num2 + start);
char *str = (char*) malloc(strlen(num1) + 2), sum;
sum = singleChar1 + singleChar - 96;
if(carryDig == 1)
sum = sum + 1;
if(start < strlen(num1)) {
if(sum >= 10) {
sum = sum - 10;
str[start] = sum + 48; //store value in each element of an array
carryDig = 1;
printf("sum of single digit is: %c\n", str[start]);
}
else if( sum < 10) {
str[start] = sum + 48; //store value in each element of an array
carryDig = 0;
printf("sum of single digit is: %c\n", str[start]);
}
add2_recur(num1,num2,start+1,carryDig);
}
else if ((start == strlen(num1)) && (carryDig ==1)){
str[start+1] = 49; // store value in each element of an array
printf("sum of single digit is: %c\n", str[start+1]);
}
return str;
}
/ ******************* /
int main（）{
char string1[20000], string2[20000], revStr1[20000], revStr2[20000], *addResult;
int digit_1, digit_2, i;
printf("Enter first number >");
fgets(string1,20000,stdin);
string1[strlen(string1)-1] = '\0';
digit_1 = digcheck(string1,0,strlen(string1)-1);
//Check if string is valid integer
if(digit_1 = 0)
printf("This number is invalid\n");
else{
printf("Enter second number >");
fgets(string2,2000,stdin);
string2[strlen(string2)-1] = '\0';
digit_2 = digcheck(string2,0,strlen(string2-1);
if(digit_2 == 0)
printf("This number is invalid\n");
else
printf("1st num is %s\n2st num is %s\n", string1, string2);
}
// reverse string
for(i=0;i<strlen(string1);i++)
revStr1[i] = string1[(strlen(string1)-1) - i];
for(i=0;i,strlen(string2);i++)
revStr2[i] = string2[(strlen(string2) -1) - i];
// compare string and pass to add2
if(strlen(revStr1) < strlen(revStr2)) {
for(i = strlen(revStr1); i < strlen(revStr2); i++)
revStr1[i] = '0';
add2(revStr1,revStr2);
}
else if(strlen(revStr2) < strlen(revStr1)) {
for(i = strlen(revStr2); i < strlen(revStr1); i++)
revStr2[i] = '0';
addResult = add2(revStr1,revStr2);
}
else
addResult = add2(revStr1,revStr2);
// print out
printf("sum is: %s\n", addResult);
return 0;
答案 2 :(得分：0)
在C中，通常不通过return实际字符串来实现这样的事情。相反，您可以使用指向传递给您的缓冲区的指针。请使用实际返回值来报告状态消息。
为了不破坏你的实际任务，让我们定义一个简单的递归函数，它将返回一个字符串，其中包含所有非字母数字字符：
#include <stdio.h>
#include <string.h>
int strip_stuff_rec(const char *input, char *output, unsigned int offset_input, unsigned offset_output) {
// Retrieve the character and move the offset
const char c = input[offset_input++];
if (c == '\0') { // Terminator; we're done!
// Terminate the output string
output[offset_output] = '\0';
return 1; // Signal success
}
// Character is alphanumeric?
if (isalnum(c)) {
// Append the character to our result and move the offset
output[offset_output++] = c;
}
// To have an error case, let's just pretend the string must not include @!
if (c == '@') {
return 0; // Signal an error
}
// Now handle the next position
return strip_stuff_rec(input, output, offset_input, offset_output);
}
int strip_stuff(const char *input, char *output) {
// Reset the output
output[0] = '\0';
// Start the recursive calls
return strip_stuff_rec(input, output, 0, 0);
}
int main(int argc, char **argv) {
// First let's set some input string
const char *input = "Hello World! -- I've had a wonderful day!";
// And we'll need a buffer for our result
char result[256];
// Now call the function and check the return value to determine
// whether it's been successful.
if (strip_stuff(input, result) == 0) {
printf("Some error happened!\n");
}
else {
printf("The stripped string is '%s'.\n", result);
}
}
答案 3 :(得分：0)
由于传递整个数组不是很理想，因此C通常在大多数表达式中将其转换为指针。
传递整个数组的一种方法是将其包含在结构中。 （虽然不是一个好的解决方案）
typedef struct
{
char s[128];
}MYSTR;