我正在尝试将字符串中的每个字母转换为它的ASCII码.运用
int letter = (atoi(ptext[i]));
给我这个错误:
error: incompatible integer to pointer conversion
passing 'char' to parameter of type 'const char *'; take the
address with & [-Werror]
int letter = (atoi(ptext[i]));
^~~~~~~~
&
/usr/include/stdlib.h:148:32: note: passing argument to parameter
'__nptr' here
extern int atoi (__const char *__nptr)
以下是我的其他一些可能相关的代码:
#include <cs50.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <ctype.h>
int main(int argc, string argv[])
{
printf("Give a string to cipher:\n");
string ptext = GetString();
int i = 0;
if (isupper(ptext[i]))
{ …我有一项任务,我应该为我的计算机科学MOOC CS50.在其中我必须通过哈佛网站上交作业,但它不接受我的代码,而它说"警告:隐含声明......"
有没有办法关闭它？
我有我使用,两种功能islower(),并isupper()和他们是什么原因造成的挂断.
我的代码似乎工作正常,它编译和一切.顺便说一句,如果有人想告诉我我的代码有多糟糕,那将是值得赞赏的.我没有得到很多(或任何)批评在网上上课.
#include <stdio.h>
#include "cs50.h"
#include <stdio.h>
#include <string.h>
int main(int argc, string argv[])
{
int salt, cipherNum;
char cipher[40];
char letter;
//// Greeting
printf("Please enter the text to ceez...\n");
//// grab text
string txxt = GetString();
if (argc == 2) // must have command line argument
{
salt = atoi(argv[1]) % 26;
//printf("Salt: %d\n", salt);
}
else // yell at user if command line arg is not there
{
printf("Not cool! …在我的代码中,我使用了贪婪的算法,以便使用最少的硬币数量.例如:我必须返回0.41美元,我可以使用的最小硬币数量是4:
1 - 0,25;
1 - 0.10;
1 - 0.05;
1 - 0.01;
有4种类型的硬币:0.25,0.10,0.05,0.01.
这是我的代码:
#include <stdio.h>
#include <cs50.h>
int main(void)
{
printf("Enter the sum, that you want to return you:");
float sum = GetFloat();
float quaters = 0.25;
float dime = 0.10;
float nickel = 0.05;
float penny = 0.01;
int count_q = 0,count_d = 0,count_n = 0,count_p = 0;
while(sum<0){
printf("Incorrect, enter the positive float number");
sum = GetFloat();
}
while(sum > 0){
if(sum - quaters >=0){
sum …当我尝试声明任何类型的变量并为其赋值时,编译器会抛出"未使用的变量错误".下面我使用'float'作为变量类型并尝试将其指定为1.5.
#include <stdio.h>
#include <cs50.h>
int main(void)
{
printf("How long is your shower?\n");
int time = GetInt();
float flow = 1.5;
}
编译器抛出此错误:
~/workspace/pset1/ $ make water
clang -ggdb3 -O0 -std=c11 -Wall -Werror -Wshadow water.c -lcs50 -lm -o water
water.c:10:11: error: unused variable 'flow' [-Werror,-Wunused-variable]
float flow = 1.5;
^
1 error generated.
make: *** [water] Error 1
我正在为CS50做第2周的pset.当使用crypt函数时,指向任何字符串的密文的char指针总是指向我加密的最后一个东西.例如:
char password[] = "AAAA";
char toCrack[] = "AAzz";
printf("%s\n", password);
printf("%s\n", toCrack);
char *toCrackCiph = crypt(toCrack, "da");
char *passwordCiph = crypt(password, "aa");
printf("%s\n", passwordCiph);
printf("%s\n", toCrackCiph);
toCrackCiph和passwordCiph彼此相等,即使它们的字符串不相同,也不是盐.
我在某处做了一个简单的指针错误吗？
谢谢,
我正在尝试查询一个名为 results 的变量，在该变量中我查询数据库以查找标题类似于从 post 方法接收到的搜索栏输入的书籍。我正在运行的查询如下：
results = db.execute("SELECT * FROM books WHERE title LIKE (%:search%)", {"search": search}).fetchall();
通过上述查询，我收到以下错误：
sqlalchemy.exc.ProgrammingError: (psycopg2.ProgrammingError) syntax error at or near "%".
如果我删除 %，或者如果我手动提供LIKEa 参数（例如：），这将按预期工作LIKE ('%the%')，但这并没有真正返回任何结果，除非搜索与数据库中的书名之一完全相同，并且它失败了通过对参数进行硬编码来使用变量替换的目的。我还想知道在使用 SQLAlchemy 查询时是否可以使用 ILIKE 来区分大小写。
我知道我可以使用对象关系映射，并使用不同的函数，例如过滤器函数等等，但是对于这个分配，我们打算不使用 ORM 并使用简单的查询。有什么建议？
我正在编写一个简单的程序,以确保我完全理解当我通过哈佛CS50时,C中的指针是如何工作的.这是代码:
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int *a = malloc(sizeof(int));
*a = 5;
printf("the address of pointer a is: %p\n", &a);
printf("the address that a is pointing to is: %p\n", a);
printf("the contents of a are: %i\n", *a);
printf("please enter a new value for a: \n");
scanf("%i", a);
printf("the address of pointer a is: %p\n", &a);
printf("the address that a is pointing to is: %p\n", a);
printf("the contents of a are: %i\n", *a);
int *b = malloc(sizeof(int));
*b …你好，我的问题是如何计算很长字符串中最长的序列AGATC（意思AGATCAGATC是两个的序列）。
我需要使用python。我现在的代码如下所示：
pattern = re.compile(r'AGATC')
matches = pattern.finditer(text_to_search)
for match in matches:
agatc += 1
print(agatc)
将打印数字 28，但我知道 22 是正确答案。
此处分析的文本：
GCTAAATTTGTTCAGCCAGATGTAGGCTTACAAATCAAGCTGTCCGCTCGGCACGGCCTACACACGTCGTGTAACTACAACAGCTAGTTAATCTGGATATCACCATGACCGAATCATAGATTTCGCCTTAAGGAGCTTTACCATGGCTTGGGATCCAATACTAAGGGCTCGACCTAGGCGAATGAGTTTCAGGTTGGCAATCAGCAACGCTCGCCATCCGGACGACGGCTTACAGTTAGTAGCATAGTACGCGATTTTCGGGAAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGAATGTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCTATCCCGTCAACTCATTCACACCGCATCCTTTCCTGCCACTGTAACTAGTCGACTGGGGAACCTCATCATCCATACTCTCCCACATTATGCCTCCCAACCTTGTTAAGCGTGGCATGCTTGGGATTGCATTGATGCTTCTTGGAGAGGACGCTTTCGTTTTGGAGATTACAGGGATCCAATTTTATCATCGGTTCGACTCCCGTAACGACTTAGCAGTAAGGGTGCTAGTTCCTGGTTAGAATCTTAATAAATCACGTCGCTTGGAGCAAGACAAAGATCGTCGTAATGCCAAGTGCACGACCACCTTCAGACTTGCAGGACCCGTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTTTTTTTCTCGATAGCTATGCGGTTCAATACAATCTTAACGCAATGCAGCGATGTGGTTTCGTACACTTAGCATAAAACCCCCCACATTAAATCGATGTACCCGCCCTCTTAGACGCCAATTTCAATGCCGAACCTCCGGCGGGTATCTCTGCACTAGGAGAAGTAGCACGTCGCTGTAGCGAACTCCTATCGTGAGATAATTTGTAGAGCTGCTCTTATAATACAATAGCTCAGATGGATTATTCCATGGACATCCCCGTGCGTTGTTTCGAGGATGGTAGGTGGAAATTTTGCCAGACCTCTAGTCTTAAACATGGTTGACGTTATAGGCGCTATCTCTTGCGTCTGGAAGTGTTAATCCGTGAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAAGAAACACGCAACTCTGGAGGAGGGCACTGCACTGCAAACTTGCGTAATATCCTTCACCCACACTTGCCTGGCCTCCTTGCTTAAAGCTCTGGCGATGCGATTTTTCGGCCCAGTAGCTGAATAGGTCATGAAATGGGCACCGAACTGGAAAGACCCATATATTCGATACTCACAACTTAATGATAGCGCGATTAAGAGCGACACCAAAAACCAAATTACGTTCACGAACCTTTGAGAGTCAAGGAGACTTAGACCGAATTGAATGATCACTGATGCGCCCGCTGATACTGAGCCTCACCATTAATCGCCGACCAATACGGCGTGTACCGGGCGCGGCCTTGCCGCATAACGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATAGATATCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTCTGTACACAGCCCCGTCCTCATTGCTAAGTGCACTGGCAACTGGACCTAAAGATTTTTCGAGTATGGCCCTCGAATCAAGCGCCCACCCAGAAACCTACGAGCCAGTAACCCCAGTAAACAAGCATTAGTGCTATATGCTTGCTGCCCACTAGGACCCTTATGGTTCATACCAGGGTGACGTGTCTTGCGGGCCAAGGATGAACCAGAAGCAAGATCCTTAGATGGACGACTGTCTCATTGCTTAAACTCCACATACCAAAGGGCGCGGTAAACGATAGTTTTAGGTAATGTTAGTCGGATGGTTGTCTGCAGCTACCAATACAGCCTGGCACCCAGGGTCTGAACAATAACGCGTGAGAGCAGCTCTCCCGCGTGTGGTGGATTTGCCGTCTATGAAATTGAGGCTCTTGCAACTATTCGCACTCGGAATGCCCTCATATCTGGTGCCTAGCGGCCTTTGCCCCGTGCCGGTAGGACTAAACTCTACGGATCGTTGACGGATCTCGATGTGGAAGATGGTTATGAAAGATAACAACGCGTGTGCTAATTGATTTAGACAAGTATTGCGGCAGTAAAAGATAATCGGCTGCAGAGTTACGAAAGACTTCCATGCATGGATTCCATTCCTTCTAGTATAGGACCCACTCTGAATACACGTCTTGCGGGCCGATCATCTCCACCGCTGCGGAAGAAAGCAATTAAGAATCTATGCTCATTAAGAGTGCGACTATAATGCGGATCTTACAGTGCTAATGATCAGGACGTCGTCCAAGCAGGCTGCATGCCGAATTTAGCTTACGTCAGGATCAGGCGTTATAGCCTGGGAATCGGACTATGAGGACGCCACGACCTCTGGGAGAAAGCTATATACATTGAGGATCGCGCCATCTTTATGAGACTCAAATGAATCTAGATAGGTAGCATTGCGGACTTGAGTTAGCACATCGGTATTGGAAGGTGAGGGTCCTGCCGCTCGTTCTATGTTCGGTTTATAGTATACAAATAGGTCATCCCGAACGTTGAAGTTAAACTCATGACACGTTGTCGTAATGAAACGGGCCTGTTATTAGGGATACAGACAAAAGGCACAAGCTGGCTTGCACATTAAGGCGCACTAGAGATCCTCACAACCGTTGCCCGCACGGAGGTCGTGTCTAACAGACAGTGAACCAGCCGTATTGGGGTGGATGACCTGAGCTTCTTGGGGCCTGTTGTACACCGCGTGTGGTTCAACTGGTACACATACTACGAATATTCGAAATCATTGTACTGTGCTCTTCGGTGCTACTGACTGTGAGCGAATGCATCCCAATCCCAAACAATGCTTGTGGTAGGAGAATTGAAACTCTCGAAGCCTGGCCCAATGTCATCTACTTTTAACATGTCGGGCCAGGAGTTACGGGCATTGCTTACTTACTTTGCCCCCTTACACCACAGCAGCGCGATTCTTGTTGTAGTAGATTTTATACGACTCGCGAATTAAATGGAACTTGTCTGTCCCATATCGATCGTGTCCATCGTAAGATGAGATTGTAGGAGCATTCGGAAGTCTATGCGGCCCAGGGACTACTACGTTAAATCTGGTCAGACGTGGTTTACAAGGCGTCCCGATCTTCTCAGAACATATGGGAAAGCACTACCGTTCCTTCACGCATACAGTTGTTCGTGCCGAACGAGTAAGCTTGCGACCAGCCCACCCGCTAGGGCTATGCAGCGGGTCATGGCTGGCGCCATACTGTGCGGACAACCCACGCTCTGGCAGAAAGCGTCTTGTGTTTTGTAGTAGCTCCAACGGTTAGACCTTCGATATCTATTCAGAGCGCGAGCGACCACTATTAGACGGCATGTAAACAATGTGTATTTGTTCGGCCCAACCGGTATATGGGTAAGACCGCGAAGGGCCTGCGCGAATACCAGCGTCCAAAAATTCCTCACCCGAGATATGCGGTTAGTACCCCTTGGGTAACGGTCCGCTACGGGTAGCGACGCGAGCCGGCCGCATCGGTTGGAGCCGAGTTGTCGGGCAGGCGAGTAACGTGTGCAATTTGATGGGCCCAAGCCTCCGGCACTATCCACCTCATACATCGACAAAAGCACCAAATATGGGGAAAAGCTGAGCGTCGATATGTACATCTACCCAGGAACCGGCCCGAACATTAGGCGGACGTGAATTTCCGACCTAGGTTCGGCTACATTTCTACGATCCAAGCACACGTGAAGGAGGAGGGGTGTTCCGACCGTAAATGAACGAGGTGCGCAGTGACCCGATGGCGTTTAGCGGATAGCCTTCCTATGCCGGCCTATGCTGTATGGTAGTTGGTTGGTGCCTCCAGAGCCACTGCACCCAATCATAGGGTCTACAGCAGCGTACTTATAAAATTGTACGGGTGACCCATATCCATTACGGGTTGCGACCAGTATAGGAGAGTATAACTGCGTGAACTAATGCGTTATGACGCTTCAGAGTTTGCTCGGGCCCGAGTTCTAGGGCTATAATGTGTTAGGGCGCAAGTATGCCAAGCTAAGATGTGGCGTGCACACTAGGAGTTGTGTTCCTCTGCAAGCAGACACGAGCACTCTGGCAGTAGTTTGACCACACCCGGGTATCACTGCTACTCCATTTCGAACAAGCTATTGGAGCGGACAAAATATGCTACTCAAGAGCATTAGTTATAGGTCTACGAGACAGAAGCAGTTACTGAGTCTGAATATTCGATATAAGTAGGCATGGAGGCGGAGCAAAACAACGTCTGCGATCAATCGTGTTGATGACGTATGGCGACTGGAAGGTAAGGACTATGGCCGGACGGAATGATTCATGTTCTGTTCAAAGCTATATTTCGAAGGGGTATATTAGCGGTCCTACACTTGGTTAGCACCCTCCCCCCTCTGGATCCTGCACTAATTCGAGCTGGCCTCCATCGGTATCAGTCCGGAAGCTCCACTCTCTATCGTAGTCCTAATCAACAGGGTGCCAGTTTGCTCACGTGGAAGTTTGAGGCCCTTTGTGCTCCATAGCCAATCACTAACCATGCACGCGCGACCCACTCTACGTCCAGATCGGCTATAATAGTTGCGCCCGGGACTGGCAGAGTAGACATGTAAGCTAGATAGAGCCCCGACATCGGCCAAGAGATCCTACGCTGCTTCCAGATAATGAGAGACATTCTAGCATTAGACATGCAAGTCGGCAGGGACTCCCCTTATCTAGTAATTTCGATGAATTGGTTTTTCGGCTAGCATCTAGTCTAGTCTAGTCTAGTCTAGTCTAGTCTAGTCTAGTCTAGTCTAGTCTAGTCTAGACCATGCCGACCTCATCATAGAAGGAATGCTCTAAACTTAGAGTGCTACTAGGAAAACTATTAATCAATGATCGTCCTGCTTACATAGCTGGACGGCGAAAGTTCTTATACTGCGGAGGTTGCTGACGTAGAGTGCGCTGGGTACAGCGGATAAGTTGATCAGGGTGGGGATAGGGTGGCTCACCGTTTATACTCATATAGATTCCTGGCGTCGACGCTGTGACAGGGTCGAGATCGAGGGGGAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGATCAGCGGAGCGGAGGGAAAATTATCACCAGAGGGTAGGGGCTCGCGACATTCTATTCAATGCATTTCAAGCTACTTACGTATTTCGGCACAGTGACTACTGCCTGCGCGGCAGCCGTAAGGTTTCCCGTCAATAGGTGGCACGTATCATTGATGAAAGTGTCAGCTAATCATTCAGGCCTTA
我正在尝试创建一个程序来检查命令行参数字符串中的重复字符。该字符串假设仅包含 26 个字符，并且所有字符都必须是字母。但是，字符串中不能有任何重复的字符，每个字母字符只能出现一次。我弄清楚了程序的前两个部分，但我不知道如何检查重复的字符。我真的可以使用一些帮助和解释。
#include <stdio.h>
#include <cs50.h>
#include <string.h>
#include <ctype.h>
int main (int argc, string argv[])
{
if (argc != 2)
{
printf("Usage: ./substitution key\n");
return 1;
}
else
{
int len = strlen(argv[1]);
if (len != 26)
{
printf("Key must contain 26 characters.\n");
return 1;
}
else
{
for (int i = 0; i < len; i++)
{
if (!isalpha(argv[1][i]))
{
printf("Usage: ./substitution key\n");
return 1;
}
}
}
}
}
所以我在 5 天前开始编程。我正在学习 cs50 课程。有一项任务（参见https://cs50.harvard.edu/x/2020/psets/2/readability/）来制作一个评估文本等级的程序。我做到了。
#include <stdio.h>
#include <cs50.h>
#include <string.h>
#include <ctype.h>
#include <math.h>
int main(void) {
int ln = 0;
int wn = 1;
int sn = 0;
string text = get_string("write your text:\n");
int l = strlen(text);
for (int i = 0; i < l; i++) {
if (isalpha(text[i])) {
ln++;
}
if ((char) (text[i]) == (char) (' ')) {
wn++;
}
if ((char) (text[i]) == (char) ('.') | (char) (text[i]) == ('!')
| (char) …cs50 ×10
c ×8
python ×2
crypt ×1
dna-sequence ×1
greedy ×1
if-statement ×1
pointers ×1
postgresql ×1
regex ×1
sqlalchemy ×1
variables ×1