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[英]Replacing a substring in a string with another substring in c using array of pointers to string
[英]C: Replacing a substring within a string using loops
我正在努力解决在字符串中替换子字符串的概念。 此特定练习不希望您使用<string.h>
或<strings.h>
内置函数。
鉴于由以下两行组成的字符串:
"Mr. Fay, is this going to be a battle of wits?"
"If it is," was the indifferent retort, "you have come unarmed!"
我必须用另一个字符串替换一个子字符串。
这是我到目前为止所拥有的,我无法将子字符串复制到新数组,并用新字符串替换子字符串:
#include <stdio.h>
#include <string.h>
int dynamic();
int main()
{
char str[]="\n\"Mr. Fay, is this going to be a battle of wits?\" \n\"If it is,\" was the indifferent retort, \"you have come unarmed!\"";
int i, j=0, k=0, l=0, n=0;
unsigned int e = n-2;
char data[150];
char newData[150];
char newStr[150];
printf("Give me a substring from the string");
gets(data);
printf("Give me a substring to replace it with");
gets(newData);
dynamic();
for (i=0; str[i] != '\0'; i++)
{
if (str[i] != data[j])
{
newStr[l] = str[i];
l++;
}
else if ((str[i+e] == data[j+e]) && (j<n))
{
newStr[l] = newData[j];
j++;
l++;
e--;
}
else if ((str[i+e] == data[j+e]) && (j>=n))
{
j++;
e--;
}
else
{
newStr[l] = str[i];
l++;
}
}
printf("original string is-");
for (k=0; k<n; k++)
printf("%c",str[k]);
printf("\n");
printf("modified string is-");
for(k=0; k<n; k++)
printf("%c",newStr[k]);
printf("\n");
}
int dynamic()
{
char str[]="\n\"Mr. Fay, is this going to be a battle of wits?\" \n\"If it is,\" was the indifferent retort, \"you have come unarmed!\"";
int i, n=0;
for (i=0; str[i] != '\0'; i++)
{
n++;
}
printf("the number of characters is %d\n",n);
return (n);
}
我试过你的问题并得到了我的代码的输出。 这是代码-编辑-这是编辑的主代码
#include <stdio.h>
#include <string.h>
int var(char *); //function declaration. I am telling CPU that I will be using this function in the later stage with one argument of type char *
int main() //main function
{
char *str="\n\"Mr. Fay, is this going to be a battle of wits?\" \n\"If it is,\" was the indifferent retort, \"you have come unarmed!\"";
int i,j=0,k=0,l=0;
char data[] = "indifferent";
char newData[] = "nonchalant";
char newStr[150];
//here 'n' is returned from the 'var' function and is received in form of r,r1,r2,r3.
int r=var(str); //getting the length of str from the function 'var' and storing in 'r'
int r1=var(data); //getting the length of data from the function 'var' and storing in 'r1'
int r2=var(newData); //getting the length of newData from the function and storing in 'r2'
unsigned int e=r1-2; //r1-2 because r1 is the data to be replaced. and string index starts from 0. Here r1 is of length 12. but we dont need to check last
//character because it is null character and the index starts from 0. not from 1. so, it is 0 to 11 and 11th is '\0'. so "12-"2"=10" characters to be compared.
for(i=0;str[i]!='\0';i++)
{
if(str[i]!=data[j])
{
newStr[l]=str[i];
l++;
}
else if((str[i+e]==data[j+e]) && (j<r2))
{
newStr[l]=newData[j];
j++;
l++;
e--;
}
else if((str[i+e]==data[j+e]) && (j>=r2))
{
j++;
e--;
}
else
{
newStr[l]=str[i];
l++;
}
}
int r3=var(newStr); //getting the length of str from the function and storing in 'r'
printf("original string is-");
for(k=0;k<r;k++)
printf("%c",str[k]);
printf("\n");
printf("modified string is-");
for(k=0;k<r3;k++)
printf("%c",newStr[k]);
printf("\n");
} // end of main function
// Below is the new function called 'var' to get the character length
//'var' is the function name and it has one parameter. I am returning integer. so, it is int var.
int var(char *stri)//common function to get length of strings and substrings
{
int i,n=0;
for(i=0;stri[i]!='\0';i++)
{
n++; //n holds the length of a string.
}
// printf("the number of characters is %d\n",n);
return (n); //returning this 'n' wherever the function is called.
}
让我解释一下代码的几个部分-
我尽力把代码写成文字。 如有疑问,可再问。 我很乐意提供帮助。 并且此代码不是最佳的。 使用的数值因您使用的单词/字符串而异。 当然,我可以为此编写一个通用代码(从字符串中自动获取数值)。 但是,我没有在这里写那个代码。 此代码适用于您的问题。 您可以更改一些变量,例如 'e' 和 ELSE-IF 部分,并尝试了解代码的工作原理。 玩它。
编辑-
int main()
{
char str[]="\n\"Mr. Fay, is this going to be a battle of wits?\" \n\"If it is,\" was the indifferent retort, \"you have come unarmed!\"";// I took this as string. The string which u need to calculate the length, You have to pass that as the function parameter.
int i,n=0;
for(i=0;str[i]!='\0';i++)
{
n++;
}
printf("the number of characters is %d\n",n);
return (n);
}// If you execute this as a separate program, you will get the number of characters in the string. Basically, you just have to modify this code to act as a separate function and when calling the function, you have to pass correct arguments.
//在函数中使用指针传递参数。
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