C language two-dimensional arrays, character array-CodePudding

# include
# include
Int main () {
Int n, m;
Char a, [100].
The scanf (" % d ", & amp; N);
Gets (a);
Char b [1000] [n].
Char c [100];
Int len=strlen (a);
Int h=0, g=0;
for(int i=0; iC=[h] a [I];
H++;
If (h==n) {
H=0;
[g] strcpy (b, c);

//printf (" % s \ n ", [g] b);
G + +;
}

}
Printf (" % s \ n ", "b" [0]).

return 0;
}
I want to write an input a string, an integer, the string of per n a "group", is I use a two-dimensional array storage per n characters each line, form a string, such as the input string abcdefgh with an integer 4. Corresponding assigned into a two-dimensional array is s [0]="abcd", s [1]="efgh", (s) as a two-dimensional array
Part of the code above comment out//printf (" % s \ n ", b [g]); If not be comments, can be normal output, if the code above the final output again, can appear problem, why ah!

CodePudding user response:

For '\ 0' terminator character array is the string; Array b, c is not a string, use strcpy, % s the operating problems,

CodePudding user response:

Char b [1000] [n + 1);
.
[g] strcpy (b, c);
Strcpyt (& amp; (b [g] [n]), '\ 0');
Seems to be a little small problems, oneself to achieve these functions, not difficult also

CodePudding user response:

 # include 
#include  
# include 

# define STR_LEN 128 

Int main () 
{
Int I, j, n, CNT; 
Int len, row, col. 
Char STR [STR_LEN]; 
Char * * PSTR; 

The fgets (STR, STR_LEN stdin); 
The scanf (" % d ", & amp; N); 
If (n & gt;=STR_LEN - 1) 
N=STR_LEN - 1; 

Len=strlen (STR); 
If (len & lt; STR_LEN) 
Len -=1; 
The row=len/n + 1; 
Col=n + 1; 
PSTR=(char * *) malloc (sizeof (char *) * row + 1); 
if (! PSTR) 
exit(0); 

for (i=0; I & lt; The row. I++) {
PSTR=[I] (char *) malloc (sizeof (char) * col); 
if (! PSTR [I]) 
exit(0); 
} 

i=0; 
CNT=0; 
While (I & lt; Len & amp; & STR [I]) {
For (j=0; J & lt; n; J++) 
PSTR [CNT] [j]=STR [I + j]; 
PSTR [CNT] [j]=0; 
I +=n; 
cnt++; 
} 

for (i=0; I & lt; CNT. I++) 
Puts (PSTR [I]); 

/* 
Int n, m; 
Char a, [100]. 
The scanf (" % d ", & amp; N); 
Gets (a); 
B//char [1000] [n]. 
B//char [1000] [n]. 
Char b [100] [100]. 
Char c [100]; 
Int len=strlen (a); 
Int h=0, g=0; 
for(int i=0; iC=[h] a [I]; 
H++; 
If (h==n) {
H=0; 
[g] strcpy (b, c); 

//printf (" % s \ n ", [g] b); 
G + +; 
} 

} 
Printf (" % s \ n ", "b" [0]). 
*/
return 0; 
}

For your reference ~

Call strcpy is the premise of c is a string that end with '\ 0',
Suggest know the strcpy,
In addition, n is the input, so you can't do the column number of the array, suggest using dynamic application space, can save a lot of space,

CodePudding user response:

reference 1st floor qiu_shaofeng response:

'\ 0' as the terminator character array is the string; Array b, c is not a string, use strcpy, % s, these operations problems

Why do they not string

CodePudding user response:

reference 3 building self-confidence boy reply:

 # include 
#include  
# include 

# define STR_LEN 128 

Int main () 
{
Int I, j, n, CNT; 
Int len, row, col. 
Char STR [STR_LEN]; 
Char * * PSTR; 

The fgets (STR, STR_LEN stdin); 
The scanf (" % d ", & amp; N); 
If (n & gt;=STR_LEN - 1) 
N=STR_LEN - 1; 

Len=strlen (STR); 
If (len & lt; STR_LEN) 
Len -=1; 
The row=len/n + 1; 
Col=n + 1; 
PSTR=(char * *) malloc (sizeof (char *) * row + 1); 
if (! PSTR) 
exit(0); 

for (i=0; I & lt; The row. I++) {
PSTR=[I] (char *) malloc (sizeof (char) * col); 
if (! PSTR [I]) 
exit(0); 
} 

i=0; 
CNT=0; 
While (I & lt; Len & amp; & STR [I]) {
For (j=0; J & lt; n; J++) 
PSTR [CNT] [j]=STR [I + j]; 
PSTR [CNT] [j]=0; 
I +=n; 
cnt++; 
} 

for (i=0; I & lt; CNT. I++) 
Puts (PSTR [I]); 

/* 
Int n, m; 
Char a, [100]. 
The scanf (" % d ", & amp; N); 
Gets (a); 
B//char [1000] [n]. 
B//char [1000] [n]. 
Char b [100] [100]. 
Char c [100]; 
Int len=strlen (a); 
Int h=0, g=0; 
for(int i=0; iC=[h] a [I]; 
H++; 
If (h==n) {
H=0; 
[g] strcpy (b, c); 

//printf (" % s \ n ", [g] b); 
G + +; 
} 

} 
Printf (" % s \ n ", "b" [0]). 
*/
return 0; 
}

Thank you, a little hard to understand the novice

CodePudding user response:

refer to the second floor qq_1457346882 response:

char b [1000] [n + 1);
.
[g] strcpy (b, c);
Strcpyt (& amp; (b [g] [n]), '\ 0');
Seems to be a little small problems, oneself to achieve these functions, also it is not hard to

I put the b [1000] [n] n to n + 1, seem to ha

CodePudding user response:

reference 5 floor Excel_lent reply:

Quote: refer to the third floor confident boy reply:

 # include 
#include  
# include 

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